Chapter 5 Understanding Quadrilaterals & Practical Geometry (Class 8 - Maths NCERT Exemplar Solutions)

To Find:

The formula for the number of diagonals in a polygon of $n$ sides.

Solution:

In a polygon with $n$ vertices, we can draw lines connecting any two vertices.

Among these lines, $n$ lines are the sides of the polygon. The remaining lines are the diagonals.

Taking the LCM to simplify the expression:

Thus, the correct option is (c).

Given:

The angles of a quadrilateral $ABCD$ are in the ratio $3 : 7 : 6 : 4$.

Solution:

Let the angles of the quadrilateral be $3x, 7x, 6x,$ and $4x$.

We know that the sum of the interior angles of a quadrilateral is $360^\circ$.

Now, let's find the measure of each angle:

• $\angle A = 3 \times 18^\circ = 54^\circ$
• $\angle B = 7 \times 18^\circ = 126^\circ$
• $\angle C = 6 \times 18^\circ = 108^\circ$
• $\angle D = 4 \times 18^\circ = 72^\circ$

Checking for parallel sides:

Since the sum of consecutive interior angles is $180^\circ$, the sides $AD$ and $BC$ are parallel ($AD \parallel BC$).

A quadrilateral with at least one pair of parallel sides is called a trapezium.

Thus, the correct option is (d).

Solution:

According to the properties of quadrilaterals:

1. In a parallelogram, diagonals bisect each other.

2. In a rhombus, diagonals bisect each other at right angles ($90^\circ$).

3. In a square, diagonals also bisect each other at right angles (but a square is also a special type of rhombus).

4. In a rectangle, diagonals are equal and bisect each other, but not necessarily at right angles.

Since the given condition is that diagonals bisect each other at right angles, it must be a rhombus.

Thus, the correct option is (a).

Solution:

The sum of the interior angles of any polygon is given by the formula $(n - 2) \times 180^\circ$, where $n$ is the number of sides.

For a quadrilateral, $n = 4$.

Thus, the correct option is (c).

Given:

$ABCD$ is a square where diagonals $AC$ and $BD$ meet at point $O$.

Solution:

In a square:

• Diagonals are equal ($AC = BD$).
• Diagonals bisect each other ($OA = OC$ and $OB = OD$). Since $AC = BD$, then $OA = OB$.
• Diagonals bisect each other at right angles ($\angle AOB = 90^\circ$).

In $\Delta AOB$:

1. Two sides are equal ($OA = OB$), so it is an isosceles triangle.

2. One angle is $90^\circ$ ($\angle AOB = 90^\circ$), so it is a right-angled triangle.

Therefore, $\Delta AOB$ is an isosceles right triangle.

Thus, the correct option is (a).

Given:

In quadrilateral $ABCD$, $AB = 5 \text{ cm}$, $CD = 8 \text{ cm}$, and $\angle A + \angle D = 180^\circ$.

Solution:

When the sum of interior angles on the same side of a transversal is $180^\circ$ (supplementary), the lines are parallel.

Since $AB = 5 \text{ cm}$ and $CD = 8 \text{ cm}$, the parallel sides are of unequal length ($AB \neq CD$).

A quadrilateral with one pair of parallel sides of unequal length is a trapezium.

Thus, the correct option is (b).

Solution:

The polygon exterior angle sum theorem states that for any convex polygon, the sum of the measures of the exterior angles, taking one at each vertex in an order, is always $360^\circ$.

This rule applies regardless of whether the shape is a triangle, quadrilateral, or any other polygon.

Since the farm land is triangular, the sum of its exterior angles is $360^\circ$.

Thus, the correct option is (c).

Solution:

In geometry, polygons are named based on the number of sides they possess:

• Pentagon: $5$ sides
• Hexagon: $6$ sides
• Heptagon: $7$ sides
• Octagon: $8$ sides

Therefore, an octagon has $8$ sides.

Thus, the correct option is (b).

Solution:

A rhombus is a special type of parallelogram where all four sides are equal. One of the fundamental properties of a rhombus is that its diagonals are perpendicular bisectors of each other.

This means the angle formed at the intersection of the two diagonals is $90^\circ$.

Answer: The diagonals of a rhombus bisect each other at right (or $90^\circ$) angles.

Solution:

A diagonal is defined as a line segment connecting two non-consecutive vertices of a polygon. In a pentagon $ABCDE$:

1. The vertices adjacent (consecutive) to $A$ are $B$ and $E$. Joining $A$ to these vertices forms the sides of the pentagon.

2. The vertices not adjacent to $A$ are $C$ and $D$.

Therefore, to form diagonals starting from vertex $A$, it must be connected to the opposite vertices.

Answer: For getting diagonals through vertex A of a pentagon ABCDE, A is joined to C and D.

Solution:

To construct a unique and specific quadrilateral, we need exactly five independent measurements. These can be various combinations such as:

• Four sides and one diagonal.
• Three sides and two diagonals.
• Four sides and one angle.
• Three sides and two included angles.
• Two adjacent sides and three angles.

Answer: For constructing a unique quadrilateral at least five measurements are required.

Solution:

We know that if the diagonals of a quadrilateral bisect each other, it is a parallelogram. Furthermore, if those diagonals bisect each other at a right angle ($90^\circ$), the parallelogram must be a rhombus.

Answer: If diagonals of a quadrilateral bisect at right angles it is a rhombus.

Solution:

There are specific quadrilaterals where the diagonals meet at $90^\circ$. These include the rhombus, the square, and the kite. However, in a general context regarding standard quadrilateral properties, the rhombus is the most common answer.

Answer: The diagonals of a rhombus (or kite/square) intersect at right angles.

Solution:

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. In a rectangle, the opposite sides are always parallel and equal, and all internal angles are $90^\circ$. Since it satisfies all the properties of a parallelogram, the statement is correct.

Answer: True (T)

Solution:

A kite is a quadrilateral that has two distinct pairs of equal adjacent sides. In a rhombus, all four sides are equal. This means the adjacent sides are equal by definition. Thus, every rhombus fits the criteria of being a kite.

Answer: True (T)

Solution:

In the Indian Perspective (NCERT), a trapezium is defined as a quadrilateral with at least one pair of parallel sides. Since a parallelogram has two pairs of parallel sides, it automatically qualifies as a trapezium under this inclusive definition.

Answer: True (T)

Solution:

A trapezium must have at least one pair of parallel sides. A kite, by definition, has two pairs of equal adjacent sides but does not require any sides to be parallel to each other.

Answer: False (F)

Solution:

In a parallelogram, opposite sides must be parallel and equal. In a kite, only the adjacent sides are equal, and opposite sides are usually not parallel or equal.

Answer: False (F)

Solution:

The diagonals of a rectangle are equal in length and bisect each other, but they only cross at a $90^\circ$ angle if the rectangle is a square. For a standard rectangle where the length is not equal to the width, the diagonals are not perpendicular.

Answer: False (F)

Solution:

Knowing only the lengths of two adjacent sides is not enough to construct a unique parallelogram because the angle between those sides can vary, resulting in different shapes. One additional measurement, such as an internal angle or the length of a diagonal, is required.

Answer: False (F)

Solution:

A curve is called a simple closed curve if it starts and ends at the same point and does not cross itself at any point.

The given figure is a six-pointed star. While it is a closed curve, it self-intersects (crosses its own lines) at several points. Therefore, it is not a simple closed curve.

Answer: False

Solution:

A polygon is called a concave polygon if at least one of its interior angles is a reflex angle (greater than $180^\circ$). In such polygons, at least one diagonal lies outside the polygon.

The given figure has an inward "V" shape. The interior angle at that vertex is greater than $180^\circ$. Hence, it is a concave polygon.

Answer: True

Solution:

A polygon is a simple closed curve made up of only line segments. The smallest possible number of sides for a polygon is three.

Since a triangle is a simple closed curve made up of three line segments, it is the most basic type of polygon.

Answer: False (A triangle is indeed a polygon).

Given:

A quadrilateral $ABCD$ where side $AB$ is extended to $P$ and side $CD$ is extended to $Q$.

To Find:

Whether $\angle ADQ + \angle CBP = \angle A + \angle C$ is true, with reasons.

Solution:

In quadrilateral $ABCD$, we know that the sum of all interior angles is $360^\circ$.

Now, consider the angles at vertices $B$ and $D$. The angles on the straight lines $AP$ and $DQ$ form linear pairs.

Adding equations (ii) and (iii), we get:

From equation (i), we can find the value of $(\angle B + \angle D)$:

Substituting this value into equation (iv):

Conclusion: Yes, the statement $\angle ADQ + \angle CBP = \angle A + \angle C$ is True.

Given:

In a parallelogram $ABCD$, $AM \perp BD$ and $CN \perp BD$.

To Prove:

$\Delta AMD \cong \Delta CNB$

Proof:

In $\Delta AMD$ and $\Delta CNB$:

By AAS (Angle-Angle-Side) congruence criterion, the two triangles are congruent.

Conclusion:

Yes, $\Delta AMD \cong \Delta CNB$.

Given:

In quadrilateral $ABCD$, $AB = AD = 5 \text{ cm}$, $BC = CD = 7 \text{ cm}$, and diagonal $BD = 6 \text{ cm}$.

Construction Steps:

1. Draw a line segment $BD = 6 \text{ cm}$.

2. With $B$ as center and radius $5 \text{ cm}$, draw an arc above $BD$.

3. With $D$ as center and radius $5 \text{ cm}$, draw another arc intersecting the previous arc at point $A$.

4. Join $AB$ and $AD$.

5. With $B$ as center and radius $7 \text{ cm}$, draw an arc below $BD$.

6. With $D$ as center and radius $7 \text{ cm}$, draw another arc intersecting the previous arc at point $C$.

7. Join $BC$ and $CD$.

Type of Quadrilateral:

In this quadrilateral, we have two distinct pairs of equal adjacent sides:

A quadrilateral that has two pairs of equal adjacent sides is called a Kite.

To Find:

The value of $x$ in the given pentagon figure.

Solution:

1. Finding Exterior Angle $\angle 1$:

In the given figure, at one of the vertices, the interior angle is marked as a right angle ($90^\circ$). The adjacent exterior angle at this vertex be labeled as $\angle 1$.

Since interior and exterior angles at any vertex lie on a straight line, they form a linear pair:

2. Applying Exterior Angle Sum Property:

The given polygon is a convex pentagon. The sum of all exterior angles of any convex polygon is always $360^\circ$.

The five exterior angles of this pentagon are $x$, $\angle 1$, $90^\circ$, $60^\circ$, and $40^\circ$.

Substituting $\angle 1 = 90^\circ$ into the equation:

Simplifying the sum of the known angles:

Solving for $x$:

Therefore, the value of $x$ is $80^\circ$.

Given:

The ratio of two adjacent angles of a parallelogram is $4 : 5$.

To Find:

The measure of each angle.

Solution:

Let the two adjacent angles be $4k$ and $5k$.

In a parallelogram, adjacent angles are supplementary, which means their sum is $180^\circ$.

Now, we find the individual angles:

First angle = $4k = 4 \times 20^\circ = 80^\circ$

Second angle = $5k = 5 \times 20^\circ = 100^\circ$

Therefore, the measures of the adjacent angles are $80^\circ$ and $100^\circ$.

Given:

The ratio of the four angles of a quadrilateral is $3 : 4 : 5 : 6$.

To Find:

The measure of all four angles.

Solution:

Let the measures of the four angles be $3k, 4k, 5k,$ and $6k$.

We know that the sum of all interior angles of a quadrilateral is $360^\circ$.

The measures of the angles are calculated as follows:

First angle = $3 \times 20^\circ = 60^\circ$

Second angle = $4 \times 20^\circ = 80^\circ$

Third angle = $5 \times 20^\circ = 100^\circ$

Fourth angle = $6 \times 20^\circ = 120^\circ$

The four angles of the quadrilateral are $60^\circ, 80^\circ, 100^\circ,$ and $120^\circ$.

Given:

$PQRS$ is a parallelogram. $PO$ is the bisector of $\angle P$ and $QO$ is the bisector of $\angle Q$.

To Find:

The measure of $\angle POQ$.

Solution:

We know that the sum of adjacent angles of a parallelogram is $180^\circ$.

Dividing both sides by 2:

In $\Delta POQ$, using the angle sum property of a triangle:

Substituting the value from equation (i):

The measure of $\angle POQ$ is $90^\circ$.

Given:

Three angles of a quadrilateral are $50^\circ, 40^\circ$ and $123^\circ$.

To Find:

The measure of the fourth angle.

Solution:

Let the fourth angle of the quadrilateral be $x$.

According to the angle sum property of a quadrilateral, the sum of all interior angles is $360^\circ$.

Adding the given angles:

The measure of the fourth angle is $147^\circ$.

Given:

The ratio of exterior angle to interior angle of a regular polygon is $1:4$.

To Find:

The number of sides of the polygon ($n$).

Solution:

Let the exterior angle be $y$ and the interior angle be $4y$.

We know that an interior angle and its corresponding exterior angle form a linear pair.

Thus, the exterior angle of the regular polygon is $36^\circ$.

The formula for the number of sides ($n$) of a regular polygon is:

The number of sides of the regular polygon is $10$.

Given:

Each interior angle of a regular polygon is $108^\circ$.

To Find:

The number of sides ($n$).

Solution:

First, we find the measure of the exterior angle.

Now, using the exterior angle formula to find $n$:

The number of sides of the polygon is $5$.

Given:

In rhombus $PAIR$, $PA = 6 \text{ cm}$ and $\angle A = 110^\circ$.

Construction Required:

In a rhombus, all sides are equal. Therefore, $PA = AI = IR = RP = 6 \text{ cm}$.

Construction Steps:

1. Draw a line segment $PA = 6 \text{ cm}$.

2. At vertex $A$, construct an angle of $110^\circ$ using a protractor.

3. With $A$ as center and radius $6 \text{ cm}$, draw an arc on the ray of the angle to locate point $I$.

4. With $I$ as center and radius $6 \text{ cm}$, draw an arc.

5. With $P$ as center and radius $6 \text{ cm}$, draw another arc to intersect the previous arc at point $R$.

6. Join $IR$ and $RP$ to complete the rhombus $PAIR$.

Given:

In rhombus $ABCD$, side $AB$ is equal to diagonal $AC$.

To Find:

All the interior angles of the rhombus.

Solution:

Since $ABCD$ is a rhombus, all its sides are equal. Let the side length be $a$.

It is given that the diagonal is also equal to the side:

Now, consider $\Delta ABC$. All three sides are equal ($AB = BC = AC = a$). Thus, $\Delta ABC$ is an equilateral triangle.

Similarly, $\Delta ADC$ is also an equilateral triangle ($AD = DC = AC = a$), so:

In a rhombus, adjacent angles are supplementary:

Since opposite angles are equal, $\angle C = \angle A = 120^\circ$.

The angles of the rhombus are $60^\circ, 120^\circ, 60^\circ,$ and $120^\circ$.

Given:

$HOPE$ is a rectangle. Its diagonals $HP$ and $OE$ intersect at $G$.

$HG = 5x + 1$ and $EG = 4x + 19$.

To Find:

The value of $x$.

Solution:

In a rectangle, the diagonals are equal in length and bisect each other.

Since the diagonals bisect each other, their halves must also be equal.

Therefore,

Substituting the given values:

Subtracting $4x$ from both sides:

Subtracting $1$ from both sides:

The value of $x$ is $18$.

Given:

$RICE$ is a rhombus with diagonals intersecting at $O$.

$OE = 5$, $OI = x + 2$, $OC = 12$, $OR = y + x$, and side $ER = z$.

To Find:

1. Values of $x, y,$ and $z$.

2. Perimeter of the rhombus.

Solution:

1. Finding $x$:

In a rhombus, the diagonals bisect each other. Therefore, the parts of the diagonal $EI$ are equal.

2. Finding $y$:

Similarly, for diagonal $CR$:

Substituting $x = 3$:

3. Finding $z$:

In a rhombus, the diagonals bisect each other at right angles. Therefore, $\Delta EOR$ is a right-angled triangle at $O$.

Using Pythagoras Theorem in $\Delta EOR$:

4. Perimeter:

In a rhombus, all four sides are equal. Each side length is $z = 13$.

The required values are $x = 3$, $y = 9$, and $z = 13$. The perimeter of the rhombus is $52$ units.

Given:

Side of the rhombus $= 4.5 \text{ cm}$.

Diagonal of the rhombus $= 6 \text{ cm}$.

Construction Required:

Let the rhombus be $ABCD$ where side $AB = BC = CD = DA = 4.5 \text{ cm}$ and diagonal $AC = 6 \text{ cm}$.

Construction Steps:

1. Draw a line segment $AC = 6 \text{ cm}$.

2. With $A$ as center and radius $4.5 \text{ cm}$, draw an arc above $AC$ and another arc below $AC$.

3. With $C$ as center and radius $4.5 \text{ cm}$, draw two arcs intersecting the previous arcs at points $B$ (above) and $D$ (below).

4. Join $AB, BC, CD,$ and $DA$.

5. $ABCD$ is the required rhombus.

Given:

Three angles of a quadrilateral are each equal to $75^\circ$.

To Find:

The measure of the fourth angle.

Solution:

Let the measure of the fourth angle be $x$.

According to the angle sum property of a quadrilateral, the sum of all its internal angles is $360^\circ$.

Thus, the fourth angle is $135^\circ$.

The correct option is (b).

Solution:

In geometry, diagonals bisect each other in all parallelograms (including squares, rectangles, and rhombuses). In a Square, the diagonals are equal, bisect each other, and are perpendicular.

For a Kite, only one diagonal is bisected by the other. For a general Trapezium or Quadrilateral, diagonals do not necessarily bisect each other.

The correct option is (a).

Solution:

In a Rectangle (and a Square), all the interior angles are equal to $90^\circ$. In a Rhombus or Trapezium, all angles are not necessarily equal.

The correct option is (a).

Solution:

The diagonals are perpendicular to each other in a Rhombus, a Square, and a Kite. In a general Parallelogram or Rectangle, diagonals are not perpendicular unless the sides are equal.

The correct option is (b).

Solution:

The diagonals are equal in length for a Rectangle, a Square, and an Isosceles Trapezium. In a Rhombus or a general Parallelogram, diagonals are usually of different lengths.

The correct option is (d).

Solution:

A figure that has all sides congruent (equal) and all angles as right angles is a Square. From the given image (Fig 8):

• P is a Trapezium.
• Q is a Parallelogram.
• R is a Square.
• S is a Rectangle.

Figure R satisfies all the given conditions.

The correct option is (c).

Solution:

A figure that has two pairs of congruent (equal) adjacent sides is a Kite. From the given image (Fig 9):

• P is a Trapezium.
• Q is a Parallelogram.
• R is a Kite.
• S is a Rectangle.

Figure R satisfies the property of having congruent adjacent sides.

The correct option is (c).

Solution:

A quadrilateral with only one pair of parallel sides is a Trapezium. From the given image (Fig 10):

• P is a Trapezium.
• Q is a Parallelogram.
• R is a Square.
• S is a Rectangle.

Figure P satisfies the property of having only one pair of sides parallel.

The correct option is (a).

Solution:

Let us analyze each figure based on the given properties:

1. Figure Q (Square): Satisfies all three properties (all sides equal, all angles $90^\circ$, and opposite sides are parallel).

2. Figure R (Rectangle): Satisfies two properties (all angles are right angles and opposite sides are parallel).

3. Figure S (Parallelogram): Satisfies the property that opposite sides are parallel.

4. Figure P (Trapezium): In a trapezium, only one pair of opposite sides is parallel. It does not satisfy the property "Opposite sides are parallel" (which refers to both pairs in the context of parallelograms), nor does it have all sides equal or all right angles.

Therefore, figure P does not satisfy any of the given properties in the sense of the parallelogram family.

The correct option is (a).

Solution:

By definition, a trapezium is a quadrilateral with at least one pair of parallel opposite sides.

The other properties mentioned (diagonals bisecting, being equal, or being perpendicular) are specific to parallelograms, rectangles, or rhombuses, and do not define a general trapezium.

The correct option is (a).

Solution:

The fundamental property of a parallelogram is that both pairs of its opposite sides are parallel. Options (b) and (c) are properties specific to a rhombus or square, and option (d) is specific to a rectangle or square.

The correct option is (a).

Solution:

An obtuse angle is an angle greater than $90^\circ$. We know that the sum of the four interior angles of a quadrilateral is $360^\circ$.

If all four angles were obtuse (say each is $91^\circ$), their sum would be:

Since the sum cannot exceed $360^\circ$, it is impossible to have four obtuse angles. However, it is possible to have three obtuse angles and one acute angle (e.g., $100^\circ, 100^\circ, 100^\circ, 60^\circ$).

The maximum number of obtuse angles is 3.

The correct option is (c).

Solution:

To divide an $n$-sided polygon into non-overlapping triangles, we pick one vertex and draw diagonals to all other non-adjacent vertices. For example:

• A quadrilateral ($n=4$) can be divided into $4 - 2 = 2$ triangles.
• A pentagon ($n=5$) can be divided into $5 - 2 = 3$ triangles.

In general, a polygon with $n$ sides can be divided into $(n - 2)$ triangles.

The correct option is (b).

Solution:

The sum of the interior angles of a polygon with $n$ sides is given by the formula:

For a pentagon, $n = 5$:

The correct option is (c).

Solution:

For a hexagon, the number of sides $n = 6$. Using the angle sum formula:

The correct option is (d).

Given:

Adjacent angles of a parallelogram are $(5x - 5)^\circ$ and $(10x + 35)^\circ$.

Solution:

Adjacent angles of a parallelogram are supplementary, meaning their sum is $180^\circ$.

Now, we find the measures of the angles:

First angle: $5(10) - 5 = 50 - 5 = 45^\circ$

Second angle: $10(10) + 35 = 100 + 35 = 135^\circ$

Ratio of the angles:

The correct option is (a).

Solution:

Let us examine the properties of the given quadrilateral:

1. All sides are equal: This property is satisfied by a rhombus and a square.

2. Opposite angles are equal: This is a general property of all parallelograms, including rhombuses and squares.

3. Diagonals bisect each other at right angles: This is a specific property of a rhombus and a square.

Since the question does not specify that all angles must be right angles (which would make it a square), the most appropriate and general name for a quadrilateral with these properties is a rhombus.

The correct option is (a).

Solution:

Consider the properties mentioned:

1. All angles are equal: In a quadrilateral, the sum of angles is $360^\circ$. If all four angles are equal, each angle must be:

2. Opposite sides are equal: A quadrilateral with opposite sides equal and each angle measuring $90^\circ$ is the definition of a rectangle.

The correct option is (a).

Solution:

Let us verify the properties for a square:

1. All sides are equal: True for a square.

2. All angles are equal: True for a square (each is $90^\circ$).

3. Diagonals are equal: True for a square.

While a rhombus has all sides equal, its diagonals and angles are not necessarily equal. While a rectangle has all angles and diagonals equal, its sides are not necessarily all equal. Only the square satisfies all these conditions simultaneously.

The correct option is (a).

Solution:

The number of diagonals in a polygon with $n$ sides is calculated using the formula:

For a hexagon, the number of sides $n = 6$. Substituting this value into the formula:

The correct option is (a).

Solution:

In a parallelogram, opposite sides are always equal. If the adjacent sides are also equal, it implies that all four sides of the parallelogram are equal.

By definition, a parallelogram with all four sides equal is a rhombus.

The correct option is (c).

Solution:

Let us analyze the diagonal properties:

1. Diagonals bisect each other: This property ensures the quadrilateral is a parallelogram.

2. Diagonals are equal: In a parallelogram, if the diagonals are equal, the figure must be a rectangle.

While a square also has equal diagonals that bisect each other, a square requires the additional property of diagonals being perpendicular. The most general quadrilateral defined by these two specific properties is the rectangle.

The correct option is (b).

Solution:

According to the Polygon Exterior Angle Sum Theorem, the sum of the measures of the exterior angles of any convex polygon, taking one at each vertex, is always $360^\circ$.

This property holds true regardless of the number of sides. Therefore, for a triangle, quadrilateral, or any other polygon, the sum remains constant.

The correct option is (b).

Solution:

A polygon that is both equilateral (all sides equal) and equiangular (all angles equal) is known as a regular polygon.

Let us check the options:

1. Rectangle: Equiangular but not necessarily equilateral.

2. Rhombus: Equilateral but not necessarily equiangular.

3. Square: It has four equal sides and four equal angles (each $90^\circ$). Thus, it is a regular quadrilateral.

The correct option is (a).

To Find:

The quadrilateral that satisfies the properties of both a kite and a parallelogram.

Solution:

Let us evaluate the properties of the given shapes:

1. Kite: A quadrilateral with two pairs of equal adjacent sides. In a kite, the diagonals are perpendicular to each other.

2. Parallelogram: A quadrilateral with two pairs of parallel opposite sides. In a parallelogram, opposite sides are equal, and diagonals bisect each other.

3. Rhombus: A rhombus has all four sides equal. Since all sides are equal, it naturally has two pairs of equal adjacent sides (like a kite). Also, since it is a parallelogram, it has opposite sides parallel (like a parallelogram).

Thus, a rhombus possesses the properties of both a kite and a parallelogram.

The correct option is (b).

Given:

The ratio of the angles of a quadrilateral is $1 : 2 : 3 : 4$.

To Find:

The measure of the smallest angle.

Solution:

Let the angles of the quadrilateral be $x, 2x, 3x,$ and $4x$.

According to the angle sum property of a quadrilateral, the sum of all interior angles is $360^\circ$.

The smallest angle corresponds to the ratio part $1$, which is $x$.

Smallest Angle = $36^\circ$

The correct option is (c).

Given:

In trapezium $ABCD$, $AB \parallel DC$ and $\angle A = 55^\circ$.

To Find:

The measure of $\angle D$.

Solution:

In a trapezium, the angles on the same side of the transversal (adjacent angles between parallel sides) are supplementary.

Since $AB \parallel DC$, $\angle A$ and $\angle D$ are consecutive interior angles.

The correct option is (d).

Given:

Three angles of a quadrilateral are each equal to $80^\circ$.

To Find:

The measure of the fourth angle.

Solution:

Let the fourth angle be $x$.

We know that the sum of the four angles of a quadrilateral is $360^\circ$.

The correct option is (b).

Given:

Each exterior angle of a regular polygon is $45^\circ$.

To Find:

The number of sides of the regular polygon ($n$).

Solution:

The sum of all exterior angles of any regular polygon is $360^\circ$.

The correct option is (a).

Given:

In parallelogram $PQRS$, $\angle P = 60^\circ$.

To Find:

The measure of $\angle Q, \angle R,$ and $\angle S$.

Solution:

In a parallelogram:

1. Opposite angles are equal.

2. Adjacent angles are supplementary.

3. Again, opposite angles are equal.

The measures of the other three angles are $120^\circ, 60^\circ,$ and $120^\circ$.

The correct option is (b).

Given:

The ratio of two adjacent angles of a parallelogram is $2 : 3$.

To Find:

The measure of the angles.

Solution:

Let the common multiplier be $x$. So the angles are $2x$ and $3x$.

Adjacent angles in a parallelogram are supplementary.

The measures of the angles are:

First Angle = $2 \times 36^\circ = 72^\circ$

Second Angle = $3 \times 36^\circ = 108^\circ$

The correct option is (a).

Given:

$PQRS$ is a parallelogram.

Solution:

In any parallelogram, the opposite angles are equal.

In $PQRS$, $\angle P$ and $\angle R$ are opposite angles.

Subtracting $\angle R$ from both sides:

The correct option is (d).

Solution:

In a parallelogram, adjacent angles are formed by a transversal cutting two parallel lines. These are consecutive interior angles, which are supplementary.

The correct option is (a).

To Find: The measure of the obtuse angle of a parallelogram when the angle between two altitudes from that vertex is $30^\circ$.

Solution:

Let $ABCD$ be a parallelogram where $\angle D$ is the obtuse angle. Draw altitudes $DM \perp AB$ and $DN \perp BC$. We consider the quadrilateral $DMBN$.

In quadrilateral $DMBN$, the sum of all angles is $360^\circ$.

Substituting the given angle between altitudes $\angle MDN = 30^\circ$ and the right angles:

In a parallelogram, opposite angles are equal.

The correct option is (b).

Given:

1. $ABCD$ is a parallelogram with base $DC$.

2. $BDCE$ is a parallelogram with base $DC$.

3. $BC \perp BD$, which means $\angle DBC = 90^\circ$.

To Find:

The measure of $\angle BEC$.

Solution:

In parallelogram $ABCD$, the opposite sides $AD$ and $BC$ are parallel ($AD \parallel BC$). Since $AB$ is extended to $E$, $AE$ acts as a transversal line. By the property of corresponding angles:

In parallelogram $BDCE$, the opposite sides $BD$ and $EC$ are parallel ($BD \parallel EC$). Considering $BC$ as a transversal line between these parallel segments, the alternate interior angles are equal:

Given that $BC \perp BD$, we have:

Substituting this value into equation (ii):

Based on the geometry of the given figure, where $\angle BAD = 30^\circ$, we substitute this into equation (i):

Now, in $\Delta BEC$, we apply the angle sum property of a triangle:

Therefore, the measure of $\angle BEC$ is $60^\circ$.

To Find: The length of the diagonal of a rectangle with sides $10 \text{ cm}$ and $24 \text{ cm}$.

Solution:

A rectangle forms two right-angled triangles with its diagonal as the hypotenuse. Using Pythagoras theorem:

The correct option is (c).

Solution:

Let the equal adjacent angles be $x$. Sum of adjacent angles of a parallelogram is $180^\circ$.

A parallelogram with each angle equal to $90^\circ$ is a rectangle.

The correct option is (a).

Solution:

The sum of the interior angles of a quadrilateral must be exactly $360^\circ$. Let's check the options:

(a) $140^\circ + 40^\circ + 20^\circ + 160^\circ = 360^\circ$ (Correct)

(b) $270^\circ + 150^\circ + ... > 360^\circ$ (Incorrect)

(c) $40^\circ + 70^\circ + 90^\circ + 60^\circ = 260^\circ$ (Incorrect)

(d) $110^\circ + 40^\circ + 30^\circ + 180^\circ = 360^\circ$ (Incorrect, as $180^\circ$ angle cannot exist in a polygon vertex).

The correct option is (a).

Solution:

The sum of the interior angles of any polygon depends only on the number of sides $n$. For any quadrilateral (convex or concave), $n = 4$.

The correct option is (c).

Solution:

For a regular polygon, the number of sides $n = \frac{360^\circ}{\text{exterior angle}}$ must be an integer greater than or equal to $3$.

(a) $\frac{360}{22} \approx 16.36$ (Not an integer)

(b) $\frac{360}{36} = 10$

(c) $\frac{360}{45} = 8$

(d) $\frac{360}{30} = 12$

The correct option is (a).

Given:

In the given figure, $BEST$ is a rhombus where the diagonals $BS$ and $ET$ intersect at point $O$. The measure of $\angle OBE = 40^\circ$. We need to find the value of $y - x$.

Solution:

In a rhombus, the opposite sides are parallel. Therefore, $TS \parallel BE$. Considering the diagonal $BS$ as a transversal line intersecting these parallel sides, the alternate interior angles are equal.

Now, consider $\Delta OST$. If we extend the side $TO$ to $E$, then $\angle SOE$ (represented by $y$) acts as an exterior angle to $\Delta OST$ at vertex $O$.

According to the exterior angle property of a triangle, the measure of an exterior angle is equal to the sum of its two interior opposite angles.

Substituting the values $y$ for $\angle SOE$, $40^\circ$ for $\angle OST$ (since $\angle OST$ is the same as $\angle BST$), and $x$ for $\angle STO$:

To find the value of $y - x$, we rearrange the equation by subtracting $x$ from both sides:

Therefore, the value of $y - x$ is $40^\circ$. The correct option is (a).

To Find:

Identify which of the given figures (a), (b), (c), or (d) is a closed curve that is also a polygon.

Solution:

A polygon is defined as a simple closed curve made up entirely of line segments. Let us carefully observe each figure provided in Fig. 15:

1. Figure (a): This is a star-shaped figure. It is a closed curve made up entirely of line segments. While it is a concave polygon, it satisfies the criteria of being a polygon.

2. Figure (b): This figure consists of two triangles joined at a single vertex. It is not considered a simple closed curve because the boundary crosses itself at that single point.

3. Figure (c): This is an open curve. Since it does not end at the same point where it started, it cannot be a polygon.

4. Figure (d): This figure contains a curved part (a semicircle) attached to a quadrilateral. Since a polygon must be made up only of line segments, this figure is not a polygon.

Conclusion:

Based on the careful observation of the shapes, only figure (a) is a polygon.

The correct option is (a).

Solution:

For a regular polygon with $n$ sides:

(a) Each exterior angle $= \frac{360^\circ}{n}$ (True)

(b) Exterior angle $= 180^\circ - \text{interior angle}$ (True, they form a linear pair)

(c) $n = \frac{360^\circ}{\text{exterior angle}}$ (True, rearrangement of option a)

(d) Each exterior angle $= \frac{(n - 2) \times 180^\circ}{n}$ (False, this is the formula for an interior angle).

The correct option is (d).

Solution:

In a square, the diagonals are equal and bisect each other at right angles ($90^\circ$). Since $PR$ and $SQ$ are diagonals intersecting at $O$:

An angle of $90^\circ$ is called a Right angle.

The correct option is (a).

To Find: All angles of the parallelogram.

Solution:

Let the adjacent angles be $1x$ and $5x$. In a parallelogram, adjacent angles are supplementary.

The measures of the adjacent angles are:

• Angle 1 $= 30^\circ$
• Angle 2 $= 5 \times 30^\circ = 150^\circ$

Since opposite angles of a parallelogram are equal, all the angles are $30^\circ, 150^\circ, 30^\circ, 150^\circ$.

The correct option is (a).

Solution:

In the parallelogram $PQRS$, we are given adjacent sides $PQ = 4 \text{ cm}$ and $QR = 6 \text{ cm}$, with an included angle $\angle PQR = 90^\circ$.

• A parallelogram with at least one right angle is a rectangle.
• Since the adjacent sides are not equal ($4 \neq 6$), it cannot be a square.

The correct option is (b).

To Find: The type of quadrilateral $PQRS$.

Solution:

Let the angles be $1k, 3k, 7k,$ and $9k$. Sum $= 360^\circ$.

Angles are: $\angle P = 18^\circ, \angle Q = 54^\circ, \angle R = 126^\circ, \angle S = 162^\circ$.

Checking for parallel sides using co-interior angles:

Since co-interior angles sum to $180^\circ$, $PQ \parallel SR$. Similarly, $\angle P + \angle S = 18^\circ + 162^\circ = 180^\circ$.

As only one pair of opposite sides is parallel (since $P + Q \neq 180^\circ$), it is a trapezium with $PQ \parallel RS$.

The correct option is (b).

To Find: The measure of $\angle R$.

Solution:

In trapezium $PQRS$ with $PQ \parallel SR$, the angles adjacent to the non-parallel sides are supplementary.

The correct option is (a).

Given:

Each interior angle of a regular polygon is $135^\circ$.

To Find:

The number of sides ($n$) of the polygon.

Solution:

We know that an interior angle and its corresponding exterior angle form a linear pair. Therefore, we can find the exterior angle first:

The number of sides of a regular polygon is given by dividing $360^\circ$ by the measure of each exterior angle:

The polygon is a regular octagon. The correct option is (c).

Solution:

A rhombus is a parallelogram with all four sides equal. One of its unique properties is that its diagonals bisect the interior angles through which they pass. Specifically, if a diagonal bisects both pairs of opposite angles, the quadrilateral must be a rhombus (or a square, which is a special rhombus).

The correct option is (c).

Solution:

A general quadrilateral requires $5$ independent measurements for construction. However, a parallelogram has inherent properties (opposite sides are equal and parallel). Therefore, we only need $3$ independent measurements, such as:

1. Two adjacent sides and the included angle.

2. Two adjacent sides and one diagonal.

The correct option is (b).

Solution:

In a rectangle, all angles are already known to be $90^\circ$. To construct it uniquely, we only need to know the lengths of its two adjacent sides (length and breadth).

The minimum measurements required are $2$. The correct option is (c).

Solution:

In a quadrilateral, vertices are named in order (clockwise or anti-clockwise). For $H-O-P-E$:

1. Side $HO$ is opposite to side $PE$.

2. Side $OP$ is opposite to side $HE$.

Answer: (HO, PE) and (OP, HE)

Solution:

Adjacent angles are angles that share a common side. In quadrilateral $ROPE$, the pairs are:

1. $\angle R$ and $\angle O$

2. $\angle O$ and $\angle P$

3. $\angle P$ and $\angle E$

4. $\angle E$ and $\angle R$

Answer: ($\angle R, \angle O$), ($\angle O, \angle P$), ($\angle P, \angle E$), and ($\angle E, \angle R$)

Solution:

Opposite angles are those that do not share a common side. In quadrilateral $WXYZ$:

1. $\angle W$ is opposite to $\angle Y$.

2. $\angle X$ is opposite to $\angle Z$.

Answer: ($\angle W, \angle Y$) and ($\angle X, \angle Z$)

Solution:

Diagonals are line segments joining non-adjacent vertices. In quadrilateral $DEFG$:

1. Joining $D$ to $F$ gives diagonal DF.

2. Joining $E$ to $G$ gives diagonal EG.

Answer: DF and EG

Solution:

According to the angle sum property of a quadrilateral, the sum of all its interior angles is $360^\circ$. Similarly, the sum of all its exterior angles is also $360^\circ$.

Answer: interior angles (or exterior angles)

Given:

A regular pentagon ($n = 5$).

Solution:

The sum of the exterior angles of any convex polygon is $360^\circ$. For a regular polygon, each exterior angle is equal.

Answer: $72^\circ$

Given:

A hexagon ($n = 6$).

Solution:

The sum of the interior angles of a polygon with $n$ sides is given by the formula $(n - 2) \times 180^\circ$.

Answer: $720^\circ$

Given:

Number of sides, $n = 18$.

Solution:

Answer: $20^\circ$

Given:

Exterior angle $= 36^\circ$.

Solution:

Answer: 10

Solution:

Any closed curve made up entirely of line segments is called a polygon. Since this particular shape has interior angles greater than $180^\circ$, it is specifically a concave polygon.

Answer: polygon (or concave polygon)

Solution:

In a kite, there is one pair of opposite angles (the angles between the non-equal sides) that are equal in measure, while the other pair of opposite angles are unequal. It is not a parallelogram because its opposite sides are not parallel.

Answer: kite

Given:

A regular pentagon ($n = 5$).

Solution:

First, find the sum of all interior angles:

Since it is a regular polygon, all angles are equal:

Answer: $108^\circ$

Solution:

A polygon with three sides is a triangle. A regular triangle is one where all sides and all angles are equal.

Answer: equilateral triangle

Given:

A hexagon ($n = 6$).

Solution:

Answer: 9

Solution:

By definition, a polygon is a simple closed curve that is formed entirely by line segments. It must not contain any curved paths, and the line segments must only meet at their endpoints.

Answer: line segments

Solution:

A polygon is termed as regular if it is both equilateral (all sides have the same length) and equiangular (all interior angles have the same measure).

Answer: angles (or interior angles)

Solution:

The sum of the interior angles of a polygon with $n$ sides is given by the formula:

Since $180^\circ$ is equal to two right angles ($2 \times 90^\circ$), we can rewrite the formula as:

Answer: $(2n - 4)$

Solution:

According to the Polygon Exterior Angle Sum Theorem, the sum of the measures of the exterior angles of any convex polygon, taking one at each vertex, is always constant regardless of the number of sides.

Answer: $360^\circ$

Solution:

A regular polygon is one where all sides and all angles are equal. A quadrilateral with four equal sides and four equal angles ($90^\circ$ each) is a square.

Answer: Square

Solution:

A quadrilateral having at least one pair of parallel opposite sides is defined as a trapezium.

Answer: trapezium

Solution:

A quadrilateral that has all four sides of equal length is called a rhombus. While a square also has equal sides, the rhombus is the more general category for this property.

Answer: rhombus

Solution:

One of the fundamental properties of a rhombus is that its diagonals are perpendicular to each other. This means they intersect at an angle of $90^\circ$.

Answer: right (or $90^\circ$)

Solution:

To construct a specific and unique quadrilateral, we require five independent measurements. These could be combinations of sides, angles, and diagonals.

Answer: Five

Solution:

A quadrilateral can be uniquely determined if three of its sides and the two included angles between those sides are known. This is often referred to as the SASAS construction criteria.

Answer: two included

Solution:

By definition, a rhombus is a special type of parallelogram in which all four sides are of equal length. Alternatively, a parallelogram with adjacent sides equal is also a rhombus.

Answer: all (or adjacent)

Solution:

A polygon is concave if at least one of its interior angles is a reflex angle, meaning it measures more than $180^\circ$.

Answer: at least one interior

Solution:

A diagonal is defined as a line segment that connects two vertices of a polygon that are not shared by the same side.

Answer: non-consecutive (or opposite)

Given:

Measure of each exterior angle $= 72^\circ$.

To Find:

Number of sides of the regular polygon ($n$).

Solution:

We know that the sum of all exterior angles of a regular polygon is $360^\circ$.

Answer: 5

Solution:

According to the properties of quadrilaterals, if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. This is a sufficient condition for a quadrilateral to be a parallelogram.

Answer: parallelogram

Given:

Adjacent sides of a parallelogram are $a = 9 \text{ cm}$ and $b = 5 \text{ cm}$.

To Find:

The perimeter of the parallelogram.

Solution:

The perimeter of a parallelogram is the sum of all its sides. Since opposite sides are equal, the formula is:

Answer: 28 cm

Solution:

A nonagon is a polygon characterized by having nine sides and nine vertices.

Answer: 9

Solution:

One of the defining properties of a rectangle is that its diagonals are equal in length. They also bisect each other, but the primary property regarding their length is that they are congruent.

Answer: equal

Solution:

A polygon with ten sides and ten vertices is called a decagon.

Answer: decagon

Solution:

A rectangle already has four right angles. If we add the condition that all its sides are equal (which happens when adjacent sides are equal), it becomes a square.

Answer: square

Solution:

One of the fundamental properties of a rectangle is that its diagonals are equal in length.

Since one diagonal is given as $6 \text{ cm}$, the other diagonal must also be of the same length.

Answer: $6 \text{ cm}$

Solution:

In a parallelogram, adjacent angles are the interior angles on the same side of a transversal. These angles are supplementary, meaning their sum is always $180^\circ$.

Answer: supplementary

Solution:

In a kite, the two diagonals are perpendicular, and only one of the diagonals (the one connecting the vertices where the unequal sides meet) is bisected by the other diagonal.

Answer: kite

Given:

In trapezium $ABCD$, $AB \parallel CD$ and $\angle A = 100^\circ$.

Solution:

Since $AB \parallel CD$, the angles $\angle A$ and $\angle D$ are consecutive interior angles (adjacent angles between parallel sides), and thus they are supplementary.

Answer: $80^\circ$

Solution:

The sum of all exterior angles for any convex polygon is always $360^\circ$.

The sum of interior angles of a polygon with $n$ sides is $(n - 2) \times 180^\circ$.

According to the question, these sums are equal:

A polygon with four sides is called a quadrilateral.

Answer: quadrilateral

Solution:

A trapezium is a quadrilateral with at least one pair of parallel sides. The angles are not necessarily equal. They are only equal if the trapezium is a rectangle or a square.

Answer: False (F)

Solution:

A rectangle is defined as a quadrilateral with four right angles. Since every square has four right angles, it satisfies the definition of a rectangle.

Answer: True (T)

Solution:

A kite only requires two pairs of equal adjacent sides. It does not require all sides to be equal or all angles to be $90^\circ$. Therefore, every kite is not a square.

Answer: False (F)

Solution:

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. In a rectangle, opposite sides are always parallel and equal. Thus, every rectangle is a parallelogram.

Answer: True (T)

Solution:

A rhombus is a quadrilateral with all four sides equal. For a rhombus to be a square, its angles must also be $90^\circ$. Since a rhombus can have any angle measure, not all rhombuses are squares.

Answer: False (F)

Solution:

The sum of the interior angles of a polygon is given by the formula $(n - 2) \times 180^\circ$, where $n$ is the number of sides. For a quadrilateral, $n = 4$.

Since the sum is $360^\circ$ and not $180^\circ$, the statement is false.

Answer: False (F)

Solution:

The number of diagonals in a polygon of $n$ sides is $\frac{n(n - 3)}{2}$. For a quadrilateral, $n = 4$.

Thus, a quadrilateral has exactly two diagonals.

Answer: True (T)

Solution:

For any triangle:

1. The sum of interior angles is $180^\circ$.

2. The sum of exterior angles (taken in order) for any polygon is $360^\circ$.

Since $360^\circ$ is indeed double of $180^\circ$, the statement is true.

Answer: True (T)

Solution:

A polygon is defined as a simple closed curve made up of only line segments.

The given figure is a star. While it is a closed curve made of line segments, it is not a simple curve because it self-intersects (crosses itself). Therefore, it is not considered a polygon in the standard sense.

Answer: False (F)

Solution:

A convex quadrilateral is one where all interior angles are less than $180^\circ$ and the diagonals lie entirely within the figure. A standard kite satisfies these properties.

Since a kite is a convex quadrilateral, the statement that it is not convex is false.

Answer: False (F)

Solution:

Let's check the condition where $\text{Sum of interior angles} = \text{Sum of exterior angles}$:

Since $n = 4$ corresponds to a quadrilateral, this equality holds true only for quadrilaterals.

Answer: True (T)

Solution:

Given condition: $\text{Sum of interior angles} = 2 \times \text{Sum of exterior angles}$

A polygon with $6$ sides is a hexagon. Thus, the statement is true.

Answer: True (T)

Solution:

A polygon is regular only if it is both equilateral (all sides are equal) and equiangular (all angles are equal).

A rhombus has all sides equal but its angles are not necessarily equal, so it is not a regular polygon. Thus, the statement is incomplete and false.

Answer: False (F)

Solution:

A polygon is considered regular if it is both equilateral (all sides are equal) and equiangular (all angles are equal).

While a rectangle is equiangular (all its angles are $90^\circ$), its adjacent sides are not necessarily equal. Therefore, it is not an equilateral quadrilateral. Only a square is a regular quadrilateral.

Answer: False (F)

Solution:

While the diagonals of a rectangle are indeed equal, there are other quadrilaterals that also possess equal diagonals. For example, an isosceles trapezium has equal diagonals but is not a rectangle.

Answer: False (F)

Solution:

One of the defining properties of a parallelogram is that both pairs of its opposite angles are equal. If a quadrilateral satisfies this property, it is guaranteed to be a parallelogram.

Answer: True (T)

Solution:

Let the interior angles be $1k, 2k,$ and $3k$.

So, the interior angles are $30^\circ, 60^\circ,$ and $90^\circ$.

The corresponding exterior angles (forming linear pairs) are:

• $180^\circ - 30^\circ = 150^\circ$
• $180^\circ - 60^\circ = 120^\circ$
• $180^\circ - 90^\circ = 90^\circ$

The ratio of exterior angles is $150 : 120 : 90$, which simplifies to $5 : 4 : 3$.

Answer: False (F)

Solution:

By observing the given figure:

1. It has 6 sides in total (the top-right slope, the bottom-right slope, the bottom edge, and the three segments forming the indented section on the left).

2. A polygon with 6 sides is called a hexagon.

3. It is concave because two of its interior angles is a reflex angle (greater than $180^\circ$).

Since the statement identifies it as a pentagon (5 sides) instead of a hexagon, the statement is false.

Answer: False (F)

Solution:

In a rhombus, the diagonals are always perpendicular bisectors of each other. However, they are not equal in length unless the rhombus is also a square.

Answer: False (F)

Solution:

A fundamental property of a rectangle is that its diagonals are equal in length and bisect each other.

Answer: True (T)

Solution:

In a rectangle, the diagonals bisect each other, but they only do so at right angles if the rectangle is a square. For a general rectangle, the angle at the intersection is not $90^\circ$.

Answer: False (F)

Solution:

A kite is a quadrilateral that has two distinct pairs of equal adjacent sides. A parallelogram is a quadrilateral where both pairs of opposite sides are parallel and equal.

Since a kite does not necessarily have its opposite sides parallel or equal, it does not satisfy the properties of a parallelogram.

Answer: False (F)

Solution:

A trapezium is a quadrilateral with at least one pair of parallel sides. A parallelogram must have two pairs of parallel sides. A trapezium with only one pair of parallel sides cannot be a parallelogram.

Answer: False (F)

Solution:

A parallelogram only requires that its opposite sides are parallel and equal. A rectangle is a special type of parallelogram where every internal angle must be a right angle ($90^\circ$). Since a general parallelogram can have angles other than $90^\circ$, every parallelogram is not a rectangle.

Answer: False (F)

Solution:

A trapezium is any quadrilateral with a pair of parallel sides. A rectangle requires two pairs of parallel sides and four right angles. Most trapeziums do not meet these strict requirements.

Answer: False (F)

Solution:

A trapezium is defined as a quadrilateral having a pair of parallel sides. Since a rectangle has two pairs of parallel sides, it also possesses at least one pair of parallel sides. Therefore, every rectangle is a trapezium.

Answer: True (T)

Solution:

A rhombus is a quadrilateral with all four sides equal. Since a square also has all four sides equal, it satisfies the definition of a rhombus.

Answer: True (T)

Solution:

A parallelogram is a quadrilateral with opposite sides parallel and equal. A square satisfies these properties (along with having equal sides and right angles), so every square is a parallelogram.

Answer: True (T)

Solution:

A trapezium requires a quadrilateral to have at least one pair of parallel sides. Since a square has two pairs of parallel sides, it is also a trapezium.

Answer: True (T)

Solution:

A trapezium is defined as a quadrilateral with at least one pair of parallel sides. Since a rhombus is a type of parallelogram, it has two pairs of opposite sides that are parallel. Therefore, it satisfies the condition of having at least one pair of parallel sides.

Answer: True (T)

Solution:

To construct a specific and unique quadrilateral, exactly five independent measurements are required. Knowing only the lengths of the four sides is insufficient because the angles between them can vary, resulting in infinitely many different quadrilaterals (a process known as "hinging").

Answer: False (F)

Solution:

An obtuse angle is defined as an angle greater than $90^\circ$. The sum of the four interior angles of a quadrilateral is always $360^\circ$. If all four angles were obtuse (for example, each being $91^\circ$), their sum would be:

Because the sum cannot be greater than $360^\circ$, it is impossible for all four angles to be obtuse.

Answer: False (F)

Solution:

This case provides a total of five measurements (four sides and one diagonal). Using these, we can construct the first triangle using the diagonal and two sides (SSS criterion), and then the second triangle on the other side of the diagonal using the remaining two sides. This results in a unique quadrilateral.

Answer: True (T)

Solution:

While four angles are mentioned, only three of them are independent because the fourth angle is automatically determined by the sum property ($360^\circ$). With three angles and only one side, we have only four independent measurements. Since five are needed for a unique construction, the quadrilateral cannot be uniquely determined.

Answer: False (F)

Solution:

This setup provides five measurements (four sides and one included angle). This is a standard condition for the unique construction of a quadrilateral, as it allows for the precise fixing of all vertices.

Answer: True (T)

Solution:

This provides five independent measurements ($3$ sides $+ 2$ diagonals). We can construct one triangle using two sides and one diagonal, and then locate the fourth vertex using the third side and the second diagonal. This results in a unique quadrilateral.

Answer: True (T)

Solution:

One of the core properties and defining tests for a parallelogram is that its diagonals bisect each other. If a quadrilateral possesses this property, it is mathematically guaranteed to be a parallelogram.

Answer: True (T)

Solution:

To construct a unique quadrilateral using three angles, the two sides provided must be adjacent sides (the sides included between the given angles). If the two sides are "any two sides" (for example, opposite sides), a unique quadrilateral cannot be guaranteed.

Answer: False (F)

Solution:

The diagonals of a parallelogram bisect each other. If we know the lengths of both diagonals and the angle at which they intersect, we can draw one diagonal, locate its midpoint, and draw the second diagonal through that midpoint at the specified angle. Connecting the four endpoints of the diagonals will result in a unique parallelogram.

Answer: True (T)

Solution:

A rhombus is a special quadrilateral where the diagonals are perpendicular bisectors of each other. Since the angle of intersection is always $90^\circ$, knowing just the lengths of the two diagonals provides enough information (five measurements: four equal segments and a fixed $90^\circ$ angle) to construct it uniquely.

Answer: True (T)

Given:

The lengths of the diagonals of a rhombus are $d_1 = 8 \text{ cm}$ and $d_2 = 15 \text{ cm}$.

To Find:

The length of the side of the rhombus.

Solution:

In a rhombus, the diagonals are perpendicular bisectors of each other. This means they intersect at $90^\circ$ and divide each other into two equal halves.

Let $ABCD$ be the rhombus where diagonals $AC$ and $BD$ intersect at point $O$.

Since the diagonals intersect at right angles, $\Delta AOB$ is a right-angled triangle where the side of the rhombus ($AB$) is the hypotenuse.

Using the Pythagoras Theorem in $\Delta AOB$:

The length of the side of the rhombus is $8.5 \text{ cm}$.

Given:

The ratio of two adjacent angles of a parallelogram is $1 : 3$.

Solution:

Let the common multiplier be $x$. Thus, the two adjacent angles are $1x$ and $3x$.

In a parallelogram, adjacent angles are supplementary (sum is $180^\circ$).

Now, we find the measures of the angles:

• First adjacent angle $= x = 45^\circ$
• Second adjacent angle $= 3x = 3 \times 45^\circ = 135^\circ$

Since opposite angles of a parallelogram are equal, the four angles are $45^\circ, 135^\circ, 45^\circ,$ and $135^\circ$.

The angles of the parallelogram are $45^\circ, 135^\circ, 45^\circ,$ and $135^\circ$.

Solution:

Among the given quadrilaterals, the Trapezium is the one that is different from the others.

Justification:

1. Parallelogram Family: Squares, rectangles, and rhombuses are all types of parallelograms. By definition, they all have two pairs of opposite sides that are parallel and equal.

2. Trapezium: A trapezium (in the standard sense) is a quadrilateral that has only one pair of parallel sides (or at least one, but it does not necessarily have two). It does not share the same symmetry and opposite-side properties that define the parallelogram family.

Conclusion: The Trapezium is the odd one out because it is not necessarily a parallelogram, unlike the square, rectangle, and rhombus.

Given:

In rectangle $ABCD$, the length of side $AB = 25 \text{ cm}$ and side $BC = 15 \text{ cm}$. Let the bisector of $\angle C$ intersect the side $AB$ at point $M$.

To Find:

The ratio in which point $M$ divides $AB$, which is $AM : MB$.

Solution:

In a rectangle, every interior angle is $90^\circ$. Since $CM$ is the bisector of $\angle C$:

In $\Delta BCM$, the angle at vertex $B$ is $90^\circ$ (property of a rectangle). We find the third angle $\angle BMC$ using the angle sum property of a triangle:

Since $\angle BCM = \angle BMC = 45^\circ$, $\Delta BCM$ is an isosceles triangle. In an isosceles triangle, sides opposite to equal angles are equal.

Now, we find the length of the remaining segment $AM$:

The ratio $AM : MB$ is:

Therefore, the bisector of $\angle C$ divides $AB$ in the ratio $2 : 3$.

Given:

$PQRS$ is a rectangle where $ST \perp PR$. The perpendicular $ST$ divides the right angle $\angle S$ into two parts ($\angle PST$ and $\angle TSR$) in the ratio $2 : 3$.

To Find:

The measure of $\angle TPQ$.

Solution:

In rectangle $PQRS$, $\angle S = 90^\circ$. Let the two parts of $\angle S$ be $2k$ and $3k$.

The measures of the two parts are:

Now, consider $\Delta STR$. It is given that $ST \perp PR$, so $\angle STR = 90^\circ$. Using the angle sum property of a triangle in $\Delta STR$:

In rectangle $PQRS$, opposite sides $PQ$ and $SR$ are parallel ($PQ \parallel SR$). The diagonal $PR$ acts as a transversal. By the property of alternate interior angles:

Therefore, the measure of $\angle TPQ$ is $36^\circ$.

Solution:

No, the photo frame is not a rectangle.

Justification:

According to the properties of a rectangle, the two diagonals must be equal in length. If one diagonal is longer than the other, the quadrilateral fails this fundamental property and cannot be classified as a rectangle. It is likely a general parallelogram or a rhombus.

Given:

Adjacent angles are $(2x - 4)^\circ$ and $(3x - 1)^\circ$.

Solution:

Adjacent angles of a parallelogram are supplementary.

Now, calculate the angles:

Angle 1 $= 2(37) - 4 = 74 - 4 = 70^\circ$

Angle 2 $= 3(37) - 1 = 111 - 1 = 110^\circ$

Since opposite angles of a parallelogram are equal, the four angles are $70^\circ, 110^\circ, 70^\circ,$ and $110^\circ$.

The angles are $70^\circ, 110^\circ, 70^\circ,$ and $110^\circ$.

Solution:

No, it cannot be a parallelogram.

Justification:

In a parallelogram, the diagonals must bisect each other. This means the point of intersection must divide each diagonal into two equal parts, creating a ratio of $1 : 1$. A ratio of $1 : 2$ implies that the diagonal is not bisected, so the figure cannot be a parallelogram.

Given:

Ratio of exterior angle to interior angle $= 1 : 5$.

Solution:

Let the exterior angle be $x$ and the interior angle be $5x$. These angles form a linear pair.

The exterior angle is $30^\circ$. The number of sides $n$ is:

The regular polygon has 12 sides (a dodecagon).

To Find: The shape formed by joining the endpoints of two crossing sticks of length $5 \text{ cm}$ each that bisect each other.

Solution:

In this scenario, the two sticks represent the diagonals of the quadrilateral formed by joining their endpoints.

1. Since both sticks are $5 \text{ cm}$ long, the diagonals are equal.

2. It is given that the sticks bisect each other.

According to the properties of quadrilaterals, a figure whose diagonals bisect each other is a parallelogram. Furthermore, if the diagonals of a parallelogram are equal, the figure is a rectangle.

Answer: The shape formed is a Rectangle.

To Find: The shape formed by joining the endpoints of two crossing sticks of length $7 \text{ cm}$ each that bisect each other at right angles.

Solution:

Here, the two sticks serve as the diagonals of the resulting quadrilateral.

1. Since both sticks are $7 \text{ cm}$ long, the diagonals are equal.

2. They bisect each other at right angles ($90^\circ$).

A quadrilateral whose diagonals are equal and are perpendicular bisectors of each other is a square.

Answer: The shape formed is a Square.

Given:

Perimeter of the kite-shaped playground $= 106 \text{ m}$. Length of one side $= 23 \text{ m}$.

To Find:

The lengths of the other three sides.

Solution:

A kite has two distinct pairs of equal adjacent sides. Let the lengths of the two pairs of sides be $a$ and $b$.

Since adjacent sides are equal, the sides are $23 \text{ m}, 23 \text{ m}, 30 \text{ m},$ and $30 \text{ m}$.

Answer: The lengths of the other three sides are $23 \text{ m}, 30 \text{ m},$ and $30 \text{ m}$.

Given:

1. $READ$ is a rectangle with diagonals $RA$ and $ED$ intersecting at $O$.

2. From the figure, the angle between the diagonals $\angle ROE = 60^\circ$.

To Find:

The measures of $\angle EAR$, $\angle RAD$, and $\angle ROD$.

Solution:

In a rectangle, diagonals are equal in length and bisect each other. Therefore, the distance from the intersection point $O$ to each vertex is equal:

1. Finding $\angle REO$:

In $\Delta ROE$, we have $OR = OE$. This makes $\Delta ROE$ an isosceles triangle. Since the vertex angle $\angle ROE = 60^\circ$, the base angles must be equal:

Since all three angles are $60^\circ$, $\Delta ROE$ is an equilateral triangle.

2. Finding $\angle EAR$:

In rectangle $READ$, $\angle REA = 90^\circ$. We find $\angle OEA$ as follows:

In $\Delta OEA$, since $OE = OA$, it is an isosceles triangle. Thus, $\angle OAE = \angle OEA = 30^\circ$. Since $\angle OAE$ is the same as $\angle EAR$, we have:

3. Finding $\angle RAD$:

In a rectangle, each interior angle is $90^\circ$. At vertex $A$, $\angle EAD = 90^\circ$.

4. Finding $\angle ROD$:

Since $E-O-D$ is a straight diagonal line, $\angle ROE$ and $\angle ROD$ form a linear pair.

Given:

1. $PAIR$ is a rectangle with diagonals $PI$ and $AR$ intersecting at $M$.

2. From the figure, $\angle MAI = 35^\circ$ and $\angle PAI = 90^\circ$.

To Find:

The measures of $\angle ARI$, $\angle RMI$, and $\angle PMA$.

Solution:

In a rectangle, diagonals are equal and bisect each other. Thus, $MP = MI = MA = MR$.

1. Finding $\angle ARI$:

In $\Delta ARI$, we know that $\angle AIR = 90^\circ$ (property of a rectangle). Since $M$ lies on diagonal $AR$, $\angle RAI = \angle MAI = 35^\circ$. Using the angle sum property in $\Delta ARI$:

So, $\angle ARI = 55^\circ$.

2. Finding $\angle RMI$:

In $\Delta AMI$, $MA = MI$ (halves of equal diagonals). This makes $\Delta AMI$ an isosceles triangle. Therefore:

Using the angle sum property in $\Delta AMI$:

Since $A-M-R$ is a straight line diagonal, $\angle AMI$ and $\angle RMI$ form a linear pair:

So, $\angle RMI = 70^\circ$.

3. Finding $\angle PMA$:

The angles $\angle PMA$ and $\angle RMI$ are vertically opposite angles formed by the intersection of diagonals $PI$ and $AR$.

So, $\angle PMA = 70^\circ$.

Given:

$ABCD$ is a parallelogram where $\angle A = 80^\circ$.

To Find:

The measures of $\angle B, \angle C,$ and $\angle D$.

Solution:

In a parallelogram, opposite angles are equal and adjacent angles are supplementary (their sum is $180^\circ$).

1. Finding $\angle C$: Since $\angle C$ is opposite to $\angle A$:

2. Finding $\angle B$: Since $\angle B$ is adjacent to $\angle A$:

3. Finding $\angle D$: Since $\angle D$ is opposite to $\angle B$:

Answer: $\angle B = 100^\circ$, $\angle C = 80^\circ$, and $\angle D = 100^\circ$.

Given:

Parallelogram $PQRS$ with $PS = 11 \text{ cm}, SR = 15 \text{ cm}, PO = 6 \text{ cm}$ (where $O$ is the diagonal intersection), and exterior angle at $Q = 60^\circ$.

Solution:

1. Finding Sides: In a parallelogram, opposite sides are equal.

2. Finding Angles: At vertex $Q$, the interior and exterior angles form a linear pair.

Since opposite angles are equal, $\angle S = \angle PQR = 120^\circ$.

Adjacent angles are supplementary: $\angle R + \angle PQR = 180^\circ \implies \angle R + 120^\circ = 180^\circ$, so $\angle R = 60^\circ$.

3. Finding Diagonal PR: Diagonals of a parallelogram bisect each other. Thus, $O$ is the midpoint of $PR$.

Answer: $\angle S = 120^\circ, \angle R = 60^\circ, PQ = 15 \text{ cm}, QR = 11 \text{ cm}$, and diagonal $PR = 12 \text{ cm}$.

Given:

$BEAM$ is a rhombus. Diagonals $BA$ and $EM$ intersect at point $O$. From the figure, $\angle OAM = 70^\circ$.

To Find:

The measures of $\angle AME$ and $\angle AEM$.

Solution:

In a rhombus, the diagonals are perpendicular bisectors of each other. Therefore, the angle at the intersection point $O$ is a right angle.

$\angle AOM = 90^\circ$

In $\Delta AOM$, using the angle sum property of a triangle:

$\angle OMA + \angle OAM + \angle AOM = 180^\circ$

$\angle OMA + 70^\circ + 90^\circ = 180^\circ$

$\angle OMA = 180^\circ - 160^\circ = 20^\circ$

Since $\angle OMA$ is the same as $\angle AME$ (as $O$ lies on the diagonal $EM$), we have:

$\angle AME = 20^\circ$

Now, in a rhombus, all sides are equal. Therefore, in $\Delta AEM$, side $AE = AM$. This makes $\Delta AEM$ an isosceles triangle. In an isosceles triangle, the angles opposite to equal sides are equal.

$\angle AEM = \angle AME$

Since we found $\angle AME = 20^\circ$, it follows that:

$\angle AEM = 20^\circ$

Answer: $\angle AME = 20^\circ$ and $\angle AEM = 20^\circ$.

Given:

1. $FIST$ is a parallelogram where diagonals $FS$ and $IT$ intersect at $O$.

2. $\angle FIO = 35^\circ$, $\angle OIS = 25^\circ$, and $\angle FOT = 110^\circ$.

To Find:

The measures of $\angle SFT$, $\angle OST$, and $\angle STO$.

Solution:

In a parallelogram, opposite sides are parallel. We use this property to find the required angles through alternate interior angles.

1. Finding $\angle STO$:

Since $FI \parallel TS$ and diagonal $IT$ is a transversal:

2. Finding $\angle SFT$:

Since $FT \parallel IS$ and diagonal $IT$ is a transversal:

Now, in $\Delta FOT$, we use the angle sum property of a triangle:

Since vertex $O$ lies on the diagonal line $FS$, $\angle OFT$ is the same as $\angle SFT$.

3. Finding $\angle OST$:

In $\Delta FOI$, we find $\angle FOI$ using the linear pair property with $\angle FOT$:

Now, using the angle sum property in $\Delta FOI$:

Since $FI \parallel TS$, diagonal $FS$ acts as a transversal:

Final Answer: $\angle SFT = 45^\circ, \angle OST = 75^\circ,$ and $\angle STO = 35^\circ$.

Given:

Parallelogram $YOUR$ where $\angle RUO = 120^\circ$. $OY$ is extended to $S$ and $\angle SRY = 50^\circ$.

Solution:

1. In parallelogram $YOUR$, opposite angles are equal:

2. Since $O-Y-S$ is a straight line, $\angle RYS$ and $\angle RYO$ form a linear pair:

3. In $\Delta RSY$, using the angle sum property:

Answer: The measure of $\angle YSR$ is $70^\circ$.

Given:

In kite $WEAR$, $\angle WEA = 70^\circ$ and $\angle ARW = 80^\circ$.

To Find:

The remaining two angles, i.e., $\angle RWE$ and $\angle RAE$.

Solution:

In a kite, one pair of opposite angles (those between the unequal sides) are equal.

Let $\angle RWE = \angle RAE = x$

We know that the sum of the interior angles of a quadrilateral is $360^\circ$.

In kite $WEAR$:

$\angle WEA + \angle ARW + \angle RWE + \angle RAE = 360^\circ$

$70^\circ + 80^\circ + x + x = 360^\circ$

$150^\circ + 2x = 360^\circ$

$2x = 360^\circ - 150^\circ$

$2x = 210^\circ$

$x = \frac{210^\circ}{2}$

$x = 105^\circ$

Therefore, the remaining two angles are $\angle RWE = 105^\circ$ and $\angle RAE = 105^\circ$.

(i) Is $RE = OM$?

Yes, because $RE$ and $OM$ are opposite sides of the rectangle $MORE$.

(ii) Is $\angle MYO = \angle RXE$?

Yes, from the figure it is given that $MY \perp OE$ and $RX \perp OE$.

(iii) Is $\angle MOY = \angle REX$?

Yes, because $OM \parallel RE$ (opposite sides of a rectangle) and $OE$ is a transversal.

(iv) Is $\triangle MYO \cong \triangle RXE$?

Yes. In $\triangle MYO$ and $\triangle RXE$:

$\angle MYO = \angle RXE = 90^\circ$ (Proved above)

$\angle MOY = \angle REX$ (Proved above)

$OM = RE$ (Proved above)

Therefore, $\triangle MYO \cong \triangle RXE$ by AAS (Angle-Angle-Side) congruence criterion.

(v) Is $MY = RX$?

Yes. Since $\triangle MYO \cong \triangle RXE$, their corresponding parts must be equal.

Given:

$LOST$ is a parallelogram where $SN \perp OL$ and $SM \perp LT$. Also, $\angle MST = 40^\circ$.

Solution:

1. To find $\angle STM$:

In right-angled triangle $\triangle SMT$:

$\angle SMT = 90^\circ$ (Given $SM \perp LT$)

$\angle MST = 40^\circ$ (Given)

Using angle sum property of a triangle:

$\angle STM + \angle SMT + \angle MST = 180^\circ$

$\angle STM + 90^\circ + 40^\circ = 180^\circ$

$\angle STM + 130^\circ = 180^\circ$

$\angle STM = 180^\circ - 130^\circ$

$\angle STM = 50^\circ$

2. To find $\angle SON$:

In parallelogram $LOST$, $LO \parallel ST$ and $LS \parallel OT$.

Opposite angles of a parallelogram are equal.

$\angle SON = 50^\circ$

3. To find $\angle NSM$:

In right-angled triangle $\triangle SON$ (as part of $\triangle SOL$ where $SN \perp OL$):

$\angle SNO = 90^\circ$

$\angle SON = 50^\circ$

In $\triangle SON$: $\angle NSO + \angle SNO + \angle SON = 180^\circ$

$\angle NSO + 90^\circ + 50^\circ = 180^\circ$

$\angle NSO = 40^\circ$

Now, adjacent angles of a parallelogram are supplementary.

$\angle SOL + \angle TSO = 180^\circ$

$50^\circ + \angle TSO = 180^\circ$

$\angle TSO = 130^\circ$

We can see from the figure that $\angle TSO = \angle MST + \angle NSM + \angle NSO$.

$130^\circ = 40^\circ + \angle NSM + 40^\circ$

$130^\circ = 80^\circ + \angle NSM$

$\angle NSM = 130^\circ - 80^\circ$

$\angle NSM = 50^\circ$

Given:

In trapezium $HARE$, $ER \parallel HA$. $EP$ is the bisector of $\angle HER$ and $RP$ is the bisector of $\angle ARE$.

From the figure, $\angle REP = 25^\circ$ and $\angle PRE = 30^\circ$.

Solution:

Since $EP$ is the bisector of $\angle HER$:

$\angle HER = 2 \times \angle REP$

$\angle HER = 2 \times 25^\circ = 50^\circ$

Since $RP$ is the bisector of $\angle ARE$:

$\angle ARE = 2 \times \angle PRE$

$\angle ARE = 2 \times 30^\circ = 60^\circ$

In trapezium $HARE$, $ER \parallel HA$. Therefore, the co-interior angles are supplementary.

1. To find $\angle EHA$:

$\angle EHA + 50^\circ = 180^\circ$

$\angle EHA = 180^\circ - 50^\circ$

$\angle EHA = 130^\circ$

2. To find $\angle HAR$:

$\angle HAR + 60^\circ = 180^\circ$

$\angle HAR = 180^\circ - 60^\circ$

$\angle HAR = 120^\circ$

Given:

$MODE$ is a parallelogram. $MQ$ is the bisector of $\angle M$ and $OQ$ is the bisector of $\angle O$.

Solution:

In parallelogram $MODE$, adjacent angles are supplementary.

Dividing the whole equation by $2$:

$\frac{1}{2}\angle M + \frac{1}{2}\angle O = \frac{180^\circ}{2}$

$\frac{1}{2}\angle M + \frac{1}{2}\angle O = 90^\circ$

Since $MQ$ and $OQ$ are angle bisectors:

$\angle QMO = \frac{1}{2}\angle M$ and $\angle QOM = \frac{1}{2}\angle O$

Substituting these in the above equation:

$\angle QMO + \angle QOM = 90^\circ$

Now, in $\triangle MQO$, using the angle sum property of a triangle:

$\angle QMO + \angle QOM + \angle MQO = 180^\circ$

$90^\circ + \angle MQO = 180^\circ$

$\angle MQO = 180^\circ - 90^\circ$

$\angle MQO = 90^\circ$

Therefore, the measure of $\angle MQO$ is $90^\circ$.

Given:

1. $ATEF$ is a rectangle.

2. $EF = EB$ (indicated by the marks on the sides).

To Find:

The values of $x$ and $y$.

Solution:

In rectangle $ATEF$, every interior angle is $90^\circ$.

Now, in $\triangle EFB$:

Since two sides are equal, $\triangle EFB$ is an isosceles right-angled triangle. Therefore, the angles opposite to these sides are equal.

By the angle sum property of a triangle:

$\angle FEB + \angle EFB + \angle EBF = 180^\circ$

$90^\circ + \angle EFB + \angle EFB = 180^\circ$

$2\angle EFB = 90^\circ$

$\angle EFB = 45^\circ$

In rectangle $ATEF$, the interior angle at vertex $F$ is $90^\circ$.

From the figure, $\angle EFA = \angle EFB + y$.

$45^\circ + y = 90^\circ$

$y = 90^\circ - 45^\circ$

$y = 45^\circ$

In rectangle $ATEF$, the opposite sides $ET$ and $FA$ are parallel ($ET \parallel FA$). The line segment $FB$ acts as a transversal intersecting these parallel lines.

The angles $x$ and $y$ are co-interior angles (also known as consecutive interior angles) on the same side of the transversal.

Substituting the value of $y$:

$x + 45^\circ = 180^\circ$

$x = 180^\circ - 45^\circ$

$x = 135^\circ$

Final Answer: The value of $x$ is $135^\circ$ and the value of $y$ is $45^\circ$.

Given:

$ABDH$ and $CEFG$ are parallelograms. $\angle B = 130^\circ$ and $\angle F = 30^\circ$.

To Find:

The value of $x$.

Solution:

In parallelogram $ABDH$:

$130^\circ + \angle D = 180^\circ$

$\angle D = 180^\circ - 130^\circ = 50^\circ$

In parallelogram $CEFG$:

$\angle C = 30^\circ$

Let the point of intersection of the dotted lines be $O$. These lines form a triangle with the base $CD$. In $\triangle OCD$:

$\angle C + \angle D + x = 180^\circ$

$30^\circ + 50^\circ + x = 180^\circ$

$80^\circ + x = 180^\circ$

$x = 180^\circ - 80^\circ$

$x = 100^\circ$

Final Answer: The value of $x$ is $100^\circ$.

Given:

$ABCD$ and $PQRS$ are rhombuses. From the figure, the distances from the center (intersection of diagonals) are:

Distance to $P = 2$ units, Distance to $Q = 2$ units.

Distance from $P$ to $A = 2$ units, Distance from $Q$ to $B = 1$ unit.

To Find:

Radius of the semicircle drawn on each side of rhombus $ABCD$.

Solution:

Let $O$ be the intersection point of the diagonals. In a rhombus, diagonals bisect each other at $90^\circ$.

Length of half-diagonal $OA = OP + PA = 2 + 2 = 4$ units.

Length of half-diagonal $OB = OQ + QB = 2 + 1 = 3$ units.

In right-angled triangle $\triangle AOB$:

$AB^2 = OA^2 + OB^2$

$AB^2 = 4^2 + 3^2$

$AB^2 = 16 + 9 = 25$

$AB = \sqrt{25} = 5$ units.

The side of the rhombus $ABCD$ is $5$ units. This side acts as the diameter of the semicircle.

Radius of semicircle = $\frac{\text{Diameter}}{2} = \frac{5}{2}$

Radius = $2.5$ units.

Final Answer: The radius of the semicircle is $2.5$ units.

Given:

$ABCDE$ is a regular pentagon. $AM$ is the bisector of $\angle A$.

To Find:

$\angle AMC$.

Solution:

Each interior angle of a regular pentagon is given by:

$\text{Interior Angle} = \frac{(n-2) \times 180^\circ}{n} = \frac{(5-2) \times 180^\circ}{5}$

$\text{Interior Angle} = \frac{3 \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ$

So, $\angle A = \angle B = \angle C = \angle D = \angle E = 108^\circ$.

Since $AM$ is the bisector of $\angle A$:

$\angle MAB = \frac{108^\circ}{2} = 54^\circ$

Now, consider the quadrilateral $ABCM$. The sum of angles in a quadrilateral is $360^\circ$.

$\angle MAB + \angle B + \angle BCM + \angle AMC = 360^\circ$

$54^\circ + 108^\circ + 108^\circ + \angle AMC = 360^\circ$

$270^\circ + \angle AMC = 360^\circ$

$\angle AMC = 360^\circ - 270^\circ$

$\angle AMC = 90^\circ$

Final Answer: $\angle AMC = 90^\circ$.

Given:

1. $EFGH$ is a rectangle.

2. $J$ is the point of intersection of diagonals $EG$ and $FH$.

3. $JF = 8x + 4$

4. $EG = 24x - 8$

To Find:

The value of $x$.

Solution:

In a rectangle, the diagonals are equal in length.

Also, the diagonals of a rectangle bisect each other. This means the intersection point $J$ is the midpoint of both diagonals.

Since $FH = EG$, we can substitute $EG$ for $FH$ in the above relation:

Now, substitute the given algebraic expressions into this equation:

$8x + 4 = \frac{1}{2} (24x - 8)$

Multiply both sides by $2$ to remove the fraction (or distribute the $\frac{1}{2}$):

$8x + 4 = 12x - 4$

Rearrange the terms to solve for $x$:

$4 + 4 = 12x - 8x$

$8 = 4x$

$x = \frac{8}{4}$

$x = 2$

Final Answer: The value of $x$ is $2$.

Given:

A parallelogram with interior angles marked as $120^\circ$, $6y$ and $(5x + 10)^\circ$.

To Find:

The values of $x$ and $y$.

Solution:

In a parallelogram, opposite angles are equal in measure.

$y = \frac{120^\circ}{6}$

$y = 20$

Also, in a parallelogram, the sum of adjacent angles is $180^\circ$.

$5x + 130^\circ = 180^\circ$

$5x = 180^\circ - 130^\circ$

$5x = 50^\circ$

$x = \frac{50^\circ}{5}$

$x = 10$

Final Answer: The values are $x = 10$ and $y = 20$.

Given:

A kite with interior angles $x$, $110^\circ$, $y$ and $60^\circ$.

To Find:

The values of $x$ and $y$.

Solution:

In a kite, exactly one pair of opposite angles (the angles between the unequal sides) are equal.

By observing the figure, the angles $110^\circ$ and $y$ are the equal pair.

We know that the sum of all interior angles of a quadrilateral is $360^\circ$.

$x + 110^\circ + y + 60^\circ = 360^\circ$

Substitute the value of $y$:

$x + 110^\circ + 110^\circ + 60^\circ = 360^\circ$

$x + 280^\circ = 360^\circ$

$x = 360^\circ - 280^\circ$

$x = 80^\circ$

Final Answer: The values are $x = 80^\circ$ and $y = 110^\circ$.

Given:

In trapezium $ABCD$, side $AB \parallel DC$.

$\angle A = (x - 20)^\circ$

$\angle D = (x + 40)^\circ$

To Find:

The value of $x$.

Solution:

Since $AB \parallel DC$ and $AD$ is a transversal, the interior angles on the same side of the transversal (co-interior angles) are supplementary.

$(x - 20)^\circ + (x + 40)^\circ = 180^\circ$

$2x + 20^\circ = 180^\circ$

$2x = 180^\circ - 20^\circ$

$2x = 160^\circ$

$x = \frac{160^\circ}{2}$

$x = 80$

Final Answer: The value of $x$ is 80.

Given:

Two angles of a quadrilateral $= 75^\circ$ each.

The other two angles are equal. Let each be $z$.

To Find:

The measure of the remaining two angles and the possible names of the figure.

Solution:

The sum of the angles of a quadrilateral is $360^\circ$.

$75^\circ + 75^\circ + z + z = 360^\circ$

$150^\circ + 2z = 360^\circ$

$2z = 360^\circ - 150^\circ$

$2z = 210^\circ$

$z = \frac{210^\circ}{2}$

$z = 105^\circ$

So, the four angles of the quadrilateral are $75^\circ$, $75^\circ$, $105^\circ$, and $105^\circ$.

Possible Figures:

1. Parallelogram: If the equal angles are opposite to each other (e.g., $75^\circ$ opposite $75^\circ$ and $105^\circ$ opposite $105^\circ$), it is a parallelogram because adjacent angles are also supplementary ($75^\circ + 105^\circ = 180^\circ$).

2. Isosceles Trapezium: If the equal angles are adjacent on the same base (e.g., base angles are $75^\circ$ and $75^\circ$ and the other two are $105^\circ$ and $105^\circ$), it forms an isosceles trapezium.

3. Kite: If only one pair of opposite angles is equal (e.g., $105^\circ$ and $105^\circ$ are between unequal sides), it could be a kite.

Final Answer: The measure of the two angles is $105^\circ$. Possible figures are Parallelogram, Isosceles Trapezium, or Kite.

Given:

In quadrilateral $PQRS$:

$\angle P = 50^\circ$, $\angle Q = 50^\circ$, $\angle R = 60^\circ$

To Find:

The measure of $\angle S$ and whether the quadrilateral is convex or concave.

Solution:

The sum of interior angles of a quadrilateral is $360^\circ$.

$\angle P + \angle Q + \angle R + \angle S = 360^\circ$

$50^\circ + 50^\circ + 60^\circ + \angle S = 360^\circ$

$160^\circ + \angle S = 360^\circ$

$\angle S = 360^\circ - 160^\circ$

$\angle S = 200^\circ$

A quadrilateral is called concave if at least one of its interior angles is greater than $180^\circ$ (reflex angle).

Since $\angle S = 200^\circ$, which is greater than $180^\circ$, the quadrilateral is concave.

Final Answer: $\angle S = 200^\circ$ and the quadrilateral is concave.

Given:

In a quadrilateral, both pairs of opposite angles are equal and each pair is supplementary.

Solution:

Let the quadrilateral be $ABCD$.

According to the question:

$\angle A = \angle C$ and $\angle B = \angle D$ (Opposite angles are equal)

Also, $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$ (Opposite angles are supplementary)

Since $\angle A = \angle C$ and $\angle A + \angle C = 180^\circ$:

$\angle A + \angle A = 180^\circ$

$2\angle A = 180^\circ$

$\angle A = \frac{180^\circ}{2} = 90^\circ$

So, $\angle C = 90^\circ$.

Similarly, since $\angle B = \angle D$ and $\angle B + \angle D = 180^\circ$:

$\angle B + \angle B = 180^\circ$

$2\angle B = 180^\circ$

$\angle B = \frac{180^\circ}{2} = 90^\circ$

So, $\angle D = 90^\circ$.

Final Answer: The measure of each angle is $90^\circ$. (The quadrilateral is a rectangle or a square).

Given:

The polygon is a regular octagon. Therefore, the number of sides $n = 8$.

Solution:

The measure of each interior angle of a regular polygon with $n$ sides is given by the formula:

$\text{Each Interior Angle} = \frac{(n - 2) \times 180^\circ}{n}$

For an octagon, $n = 8$:

$\text{Each Interior Angle} = \frac{(8 - 2) \times 180^\circ}{8}$

$\text{Each Interior Angle} = \frac{6 \times 180^\circ}{8}$

$\text{Each Interior Angle} = \frac{1080^\circ}{8}$

$\text{Each Interior Angle} = 135^\circ$

Final Answer: The measure of each angle of a regular octagon is $135^\circ$.

Solution:

We know that the measure of each exterior angle of a regular polygon with $n$ sides is $\frac{360^\circ}{n}$.

1. For a regular pentagon ($n = 5$):

$\text{Exterior Angle}_1 = \frac{360^\circ}{5} = 72^\circ$

2. For a regular decagon ($n = 10$):

$\text{Exterior Angle}_2 = \frac{360^\circ}{10} = 36^\circ$

Finding the Ratio:

$\text{Ratio} = \frac{\text{Exterior angle of pentagon}}{\text{Exterior angle of decagon}}$

$\text{Ratio} = \frac{72^\circ}{36^\circ}$

$\text{Ratio} = \frac{2}{1} = 2 : 1$

Final Answer: The exterior angles are $72^\circ$ and $36^\circ$ respectively, and their ratio is $2 : 1$.

Given:

A polygon with five exterior angles given as: $x^\circ$, $85^\circ$, $20^\circ$, $92^\circ$, and $89^\circ$.

Solution:

We know that the sum of the exterior angles of any convex polygon is always $360^\circ$.

$x^\circ + 85^\circ + 20^\circ + 92^\circ + 89^\circ = 360^\circ$

$x^\circ + 286^\circ = 360^\circ$

$x^\circ = 360^\circ - 286^\circ$

$x^\circ = 74^\circ$

Final Answer: The value of $x$ is 74.

Given:

Three angles of a quadrilateral are equal. Let each equal angle be $y$.

The fourth angle $= 120^\circ$.

Solution:

The sum of all interior angles of a quadrilateral is $360^\circ$.

$y + y + y + 120^\circ = 360^\circ$

$3y + 120^\circ = 360^\circ$

$3y = 360^\circ - 120^\circ$

$3y = 240^\circ$

$y = \frac{240^\circ}{3}$

$y = 80^\circ$

Final Answer: The measure of each equal angle is $80^\circ$.

Given:

In quadrilateral $HOPE$, $PS$ is the bisector of $\angle P$ and $ES$ is the bisector of $\angle E$. These bisectors meet at point $S$.

To Prove:

$\angle PSE = \frac{1}{2}(\angle H + \angle O)$

Proof:

We know that the sum of the interior angles of a quadrilateral is $360^\circ$.

In quadrilateral $HOPE$:

$\angle H + \angle O + \angle P + \angle E = 360^\circ$

$\angle P + \angle E = 360^\circ - (\angle H + \angle O)$

Multiplying the entire equation by $\frac{1}{2}$:

$\frac{1}{2} (\angle P + \angle E) = \frac{1}{2} [360^\circ - (\angle H + \angle O)]$

$\frac{1}{2} \angle P + \frac{1}{2} \angle E = 180^\circ - \frac{1}{2} (\angle H + \angle O)$

In $\triangle PSE$, by the angle sum property of a triangle:

$\angle PSE + \angle SPE + \angle SEP = 180^\circ$

Since $PS$ and $ES$ are angle bisectors, we have $\angle SPE = \frac{1}{2} \angle P$ and $\angle SEP = \frac{1}{2} \angle E$.

$\angle PSE + (\frac{1}{2} \angle P + \frac{1}{2} \angle E) = 180^\circ$

Now, substitute the value of $(\frac{1}{2} \angle P + \frac{1}{2} \angle E)$ calculated earlier:

$\angle PSE + [180^\circ - \frac{1}{2} (\angle H + \angle O)] = 180^\circ$

$\angle PSE + 180^\circ - \frac{1}{2} (\angle H + \angle O) = 180^\circ$

$\angle PSE = 180^\circ - 180^\circ + \frac{1}{2} (\angle H + \angle O)$

$\angle PSE = \frac{1}{2} (\angle H + \angle O)$

Hence Proved.

Given:

$ABCD$ is a parallelogram. Diagonals $AC$ and $BD$ intersect at $O$.

$\angle AOB = 100^\circ$ and $\angle DBC = 30^\circ$.

To Find:

The values of $x$, $y$, and $z$.

Solution:

1. To find x:

Since $AOC$ is a straight line, $\angle AOB$ and $\angle BOC$ form a linear pair.

$100^\circ + x = 180^\circ$

$x = 180^\circ - 100^\circ = 80^\circ$

2. To find y:

In $\triangle BOC$, the sum of angles is $180^\circ$.

$x + y + \angle OBC = 180^\circ$

Given $\angle DBC = 30^\circ$, which is the same as $\angle OBC$.

$80^\circ + y + 30^\circ = 180^\circ$

$y + 110^\circ = 180^\circ$

$y = 180^\circ - 110^\circ = 70^\circ$

3. To find z:

In parallelogram $ABCD$, opposite sides are parallel, so $AD \parallel BC$. $BD$ acts as a transversal.

$z = 30^\circ$

Final Answer: $x = 80^\circ$, $y = 70^\circ$, and $z = 30^\circ$.

Solution:

No, a quadrilateral with perpendicular diagonals is not always a rhombus.

For a quadrilateral to be a rhombus, the diagonals must not only be perpendicular but must also bisect each other. Additionally, all four sides must be equal.

A Kite is a prime example of a quadrilateral where the diagonals are perpendicular to each other, but it is not a rhombus (as only two pairs of adjacent sides are equal, and only one diagonal is bisected).

Justification Figure:

In the figure above, the diagonals intersect at $90^\circ$, but the sides are not all equal, hence it is not a rhombus.

Given:

Trapezium $ABCD$ with $AB \parallel CD$.

$\angle A : \angle D = 2 : 1$

$\angle B : \angle C = 7 : 5$

Solution:

Since $AB \parallel CD$, the pairs of angles $\angle A, \angle D$ and $\angle B, \angle C$ are co-interior angles, and their sum is $180^\circ$.

1. For angles A and D:

Let $\angle A = 2k$ and $\angle D = 1k$.

$2k + k = 180^\circ$

$3k = 180^\circ$

$k = 60^\circ$

$\angle A = 2 \times 60^\circ = 120^\circ$

$\angle D = 1 \times 60^\circ = 60^\circ$

2. For angles B and C:

Let $\angle B = 7m$ and $\angle C = 5m$.

$7m + 5m = 180^\circ$

$12m = 180^\circ$

$m = \frac{180^\circ}{12} = 15^\circ$

$\angle B = 7 \times 15^\circ = 105^\circ$

$\angle C = 5 \times 15^\circ = 75^\circ$

Final Answer: The angles are $\angle A = 120^\circ$, $\angle B = 105^\circ$, $\angle C = 75^\circ$, and $\angle D = 60^\circ$.

Given:

1. Line $l \parallel m$.

2. Transversal $p$ intersects $l$ at $X$ and $m$ at $Y$.

3. $XP, XQ, YP,$ and $YQ$ are the bisectors of the four interior angles formed at $X$ and $Y$.

Solution:

Yes, the quadrilateral $PXQY$ is a rectangle.

Reasoning:

Let the two interior angles at point $X$ be $\angle AX Y$ and $\angle BXY$, and the two interior angles at point $Y$ be $\angle CYX$ and $\angle DYX$.

1. To show $\angle XPY = 90^\circ$ and $\angle XQY = 90^\circ$:

We know that consecutive interior angles are supplementary.

Dividing by $2$:

$\frac{1}{2}\angle AXY + \frac{1}{2}\angle CYX = 90^\circ$

Since $XP$ and $YP$ are angle bisectors, we have $\angle PXY = \frac{1}{2}\angle AXY$ and $\angle PYX = \frac{1}{2}\angle CYX$.

$\angle PXY + \angle PYX = 90^\circ$

In $\triangle XPY$, using the angle sum property:

$\angle XPY = 180^\circ - (\angle PXY + \angle PYX) = 180^\circ - 90^\circ = 90^\circ$

Similarly, we can prove that $\angle XQY = 90^\circ$.

2. To show $\angle PXQ = 90^\circ$ and $\angle PYQ = 90^\circ$:

The two interior angles at vertex $X$ (on either side of the transversal) form a linear pair.

Dividing by $2$:

$\frac{1}{2}\angle AXY + \frac{1}{2}\angle BXY = 90^\circ$

$\angle PXY + \angle QXY = 90^\circ$

$\angle PXQ = 90^\circ$

Similarly, we can prove that $\angle PYQ = 90^\circ$.

Since all the interior angles of the quadrilateral $PXQY$ are $90^\circ$, it is a rectangle.

Given:

$ABCD$ is a parallelogram. $AX$ is the bisector of $\angle A$ and $CY$ is the bisector of $\angle C$.

To Find:

Is $AXCY$ a parallelogram?

Solution:

Yes, $AXCY$ is a parallelogram.

In parallelogram $ABCD$, opposite angles are equal.

Dividing by $2$:

$\frac{1}{2} \angle A = \frac{1}{2} \angle C$

Since $AX$ and $CY$ are bisectors:

$\angle XAY = \angle XCY$

Also, in parallelogram $ABCD$, $AB \parallel CD$, which implies $AY \parallel XC$.

Now, consider transversal $AX$ intersecting parallel lines $AB$ and $CD$:

Since $\angle XAY = \angle XCY$, we have $\angle AXD = \angle XCY$.

These are corresponding angles for lines $AX$ and $CY$ with transversal $CD$. Therefore, $AX \parallel CY$.

Since both pairs of opposite sides ($AY \parallel XC$ and $AX \parallel CY$) are parallel, $AXCY$ is a parallelogram.

Given:

$ABCD$ is a parallelogram and diagonal $AC$ bisects $\angle A$.

Solution:

Yes, it will also bisect the other angle ($\angle C$).

Reason:

Since $ABCD$ is a parallelogram, $AB \parallel DC$ and $AD \parallel BC$.

1. Since $AB \parallel DC$ and $AC$ is a transversal:

2. Since $AD \parallel BC$ and $AC$ is a transversal:

3. It is given that $AC$ bisects $\angle A$:

From the above statements, we can conclude that:

$\angle DCA = \angle BCA$

Since $\angle DCA = \angle BCA$, diagonal $AC$ also bisects $\angle C$.

Note: In such a case, the parallelogram is actually a rhombus.

Given:

In parallelogram $ABCD$, let $\angle B$ be the obtuse angle. Altitudes $BM$ and $BN$ are drawn from $B$ to sides $AD$ and $CD$ respectively. The angle between them, $\angle MBN = 45^\circ$.

Solution:

Consider the quadrilateral $BMDN$.

In this quadrilateral:

$\angle BMD = 90^\circ$ (Since $BM$ is an altitude to $AD$)

$\angle BND = 90^\circ$ (Since $BN$ is an altitude to $CD$)

$\angle MBN = 45^\circ$ (Given)

Sum of angles of a quadrilateral is $360^\circ$:

$\angle BMD + \angle BND + \angle MBN + \angle D = 360^\circ$

$90^\circ + 90^\circ + 45^\circ + \angle D = 360^\circ$

$225^\circ + \angle D = 360^\circ$

$\angle D = 360^\circ - 225^\circ = 135^\circ$

In parallelogram $ABCD$, opposite angles are equal:

$\angle B = \angle D = 135^\circ$

Adjacent angles are supplementary:

$\angle A + \angle D = 180^\circ$

$\angle A + 135^\circ = 180^\circ$

$\angle A = 180^\circ - 135^\circ = 45^\circ$

Since $\angle C = \angle A$, $\angle C = 45^\circ$.

Final Answer: The angles of the parallelogram are $45^\circ, 135^\circ, 45^\circ, 135^\circ$.

Given:

$ABCD$ is a rhombus. The perpendicular bisector of $AB$ passes through vertex $D$.

Solution:

Let $M$ be the midpoint of $AB$. Since $DM$ is the perpendicular bisector of $AB$, any point on $DM$ is equidistant from $A$ and $B$.

In a rhombus, all sides are equal. Therefore:

From the two equations above, we get:

$AD = AB = BD$

This means $\triangle ABD$ is an equilateral triangle.

In an equilateral triangle, each angle is $60^\circ$.

$\angle A = 60^\circ$

In rhombus $ABCD$:

$\angle C = \angle A = 60^\circ$ (Opposite angles)

$\angle A + \angle B = 180^\circ$ (Adjacent angles are supplementary)

$60^\circ + \angle B = 180^\circ$

$\angle B = 120^\circ$

$\angle D = \angle B = 120^\circ$ (Opposite angles)

Final Answer: The angles of the rhombus are $60^\circ, 120^\circ, 60^\circ, 120^\circ$.

Given:

1. $ABCD$ is a parallelogram with $\angle C = 45^\circ$.

2. $PRQA$ is another parallelogram where $P$ lies on $AB$ and $Q$ lies on $AD$.

To Find:

The measure of $\angle R$.

Solution:

In parallelogram $ABCD$:

Given $\angle C = 45^\circ$, therefore $\angle A = 45^\circ$.

Now, consider parallelogram $PRQA$. In this parallelogram, $\angle A$ and $\angle R$ are opposite angles.

Since $\angle A = 45^\circ$:

$\angle R = 45^\circ$

Final Answer: The measure of $\angle R$ is $45^\circ$.

Given:

1. $ABCD$ is a parallelogram where $AD \parallel BC$ and $AB \parallel DC$.

2. The bisector of $\angle A$ intersects $BC$ at $E$ such that $BE = EC = \frac{1}{2}BC$.

To Find:

Whether the angle bisector of $\angle B$ also bisects $AD$.

Solution:

Step 1: Establishing the relationship between sides AB and BC.

From the above two equations, we get:

$\angle BAE = \angle AEB$

In $\triangle ABE$, since the base angles are equal, the sides opposite to them must be equal.

Since $E$ is the midpoint of $BC$ (given that $AE$ bisects $BC$):

$BE = \frac{1}{2} BC$

Substituting $AB = BE$:

$AB = \frac{1}{2} BC$, which means $BC = 2AB$.

Step 2: Checking the bisector of $\angle B$.

Let the bisector of $\angle B$ intersect $AD$ at point $F$.

Therefore, $\angle ABF = \angle AFB$.

In $\triangle ABF$, since the angles are equal, the opposite sides are equal:

Step 3: Conclusion.

In parallelogram $ABCD$, opposite sides are equal ($AD = BC$).

From Step 1, we found $BC = 2AB$. Therefore, $AD = 2AB$.

From Step 2, we found $AF = AB$.

Comparing these:

$AF = \frac{1}{2} AD$

Since $AF$ is exactly half of $AD$, the point $F$ is the midpoint of $AD$.

Final Answer: Yes, the angle bisector of $\angle B$ will also bisect $AD$ because the sides of the parallelogram are in the ratio $1 : 2$.

Given:

1. $ABCDE$ is a regular pentagon.

2. $ABFG$ is a square formed on the opposite side of $AB$.

Solution:

The interior angle of a regular pentagon ($n = 5$) is:

$\angle ABC = \frac{(5 - 2) \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ$

The interior angle of a square is $90^\circ$. So, $\angle ABF = 90^\circ$.

Since the pentagon and square are on opposite sides of $AB$, the angle $\angle CBF$ is:

$\angle CBF = 360^\circ - (\angle ABC + \angle ABF)$

$\angle CBF = 360^\circ - (108^\circ + 90^\circ) = 360^\circ - 198^\circ = 162^\circ$

In regular pentagon $ABCDE$, all sides are equal, so $BC = AB$.

In square $ABFG$, all sides are equal, so $BF = AB$.

Therefore, $BC = BF$. This makes $\triangle BCF$ an isosceles triangle.

In $\triangle BCF$:

$\angle BCF = \angle BFC$ (Angles opposite to equal sides)

Using the angle sum property:

$\angle CBF + \angle BCF + \angle BFC = 180^\circ$

$162^\circ + 2\angle BCF = 180^\circ$

$2\angle BCF = 180^\circ - 162^\circ = 18^\circ$

$\angle BCF = \frac{18^\circ}{2} = 9^\circ$

Final Answer: The measure of $\angle BCF$ is $9^\circ$.

Solution:

Let's analyze the number of acute angles in a convex polygon with $n$ sides.

1. If an interior angle is acute ($< 90^\circ$), then its corresponding exterior angle must be obtuse ($> 90^\circ$) because they form a linear pair.

2. We know that the sum of all exterior angles of any convex polygon is exactly $360^\circ$.

3. If a polygon has 4 or more acute interior angles, it must have 4 or more obtuse exterior angles. However, the sum of 4 obtuse angles (each $> 90^\circ$) would be greater than $4 \times 90^\circ = 360^\circ$.

4. Since the sum cannot exceed $360^\circ$, a convex polygon can have at most 3 obtuse exterior angles, which means it can have at most 3 acute interior angles.

Observations:

1. Quadrilateral ($n = 4$): Maximum 3 acute angles (e.g., a kite with angles $70^\circ, 70^\circ, 70^\circ, 150^\circ$).

2. Pentagon ($n = 5$): Maximum 3 acute angles.

3. Hexagon ($n = 6$): Maximum 3 acute angles.

Generalisation:

For any convex polygon with $n$ sides ($n \geq 3$), the maximum number of acute interior angles it can have is 3.

Given:

1. $FD \parallel BC \parallel AE$

2. $AC \parallel ED$

3. In $\triangle ABC$, $\angle ABC = 64^\circ$ and $\angle BAC = 52^\circ$.

Solution:

First, find $\angle BCA$ in $\triangle ABC$:

$\angle BCA = 180^\circ - (\angle ABC + \angle BAC)$

$\angle BCA = 180^\circ - (64^\circ + 52^\circ) = 180^\circ - 116^\circ = 64^\circ$

Since $BC \parallel AE$ and $AC$ is a transversal:

Now, consider the quadrilateral $ACDE$.

Since $AE \parallel FD$ (part of the given $AE \parallel BC \parallel FD$) and $AC \parallel ED$, the quadrilateral $ACDE$ is a parallelogram.

In a parallelogram, adjacent angles are supplementary.

$64^\circ + x = 180^\circ$

$x = 180^\circ - 64^\circ$

$x = 116^\circ$

Final Answer: The value of $x$ is 116.

Given:

In trapezium $ABCD$:

$AB \parallel DC$

$AD = BC = 10 \text{ cm}$

$DC = 20 \text{ cm}$

$AB = x \text{ cm}$

$\angle DAB = 60^\circ$

To Find:

The value of $x$.

Construction Required:

Draw a line $CE$ parallel to $AD$, such that $E$ lies on $AB$.

Solution:

In quadrilateral $AECD$:

Since both pairs of opposite sides are parallel, $AECD$ is a parallelogram.

In a parallelogram, opposite sides are equal.

Now, since $AD \parallel CE$ and $AB$ is a transversal:

In $\triangle CEB$:

Since $CE = BC$, $\triangle CEB$ is an isosceles triangle. The angles opposite to equal sides must be equal.

Now, by the angle sum property in $\triangle CEB$:

$\angle ECB + \angle CEB + \angle B = 180^\circ$

$\angle ECB + 60^\circ + 60^\circ = 180^\circ$

$\angle ECB + 120^\circ = 180^\circ$

$\angle ECB = 60^\circ$

Since all three angles of $\triangle CEB$ are $60^\circ$, it is an equilateral triangle.

In an equilateral triangle, all sides are equal. Therefore:

From the figure, we can see that:

$AB = AE + EB$

$x = 20 \text{ cm} + 10 \text{ cm}$ (Using (i) and (iii))

$x = 30 \text{ cm}$

Final Answer: The value of $x$ is 30.

Given:

In trapezium $ABCD$, $AB \parallel DC$, $\angle A = 105^\circ$, $AD = 3\text{ cm}$, $AB = 4\text{ cm}$, and $CD = 8\text{ cm}$.

To Construct:

A trapezium $ABCD$ with the given measurements.

Steps of Construction:

1. Draw a line segment $AB = 4\text{ cm}$.

2. At point $A$, construct an angle $\angle XAB = 105^\circ$ using a protractor or compass.

3. With $A$ as center and radius $3\text{ cm}$, draw an arc on ray $AX$ to mark point $D$. Thus, $AD = 3\text{ cm}$.

4. Since $AB \parallel DC$, the sum of consecutive interior angles $\angle A$ and $\angle D$ must be $180^\circ$.

5. At point $D$, construct an angle of $75^\circ$ with respect to $AD$ such that the new ray $DY$ is parallel to $AB$.

6. With $D$ as center and radius $8\text{ cm}$, draw an arc on ray $DY$ to mark point $C$. Thus, $DC = 8\text{ cm}$.

7. Join points $B$ and $C$.

8. $ABCD$ is the required trapezium.

Given:

In parallelogram $ABCD$, $AB = 4\text{ cm}$, $BC = 5\text{ cm}$, and $\angle B = 60^\circ$.

To Construct:

Parallelogram $ABCD$.

Steps of Construction:

1. Draw a line segment $AB = 4\text{ cm}$.

2. At point $B$, construct an angle $\angle ABX = 60^\circ$.

3. With $B$ as center and radius $5\text{ cm}$, draw an arc on ray $BX$ to mark point $C$.

4. We know that in a parallelogram, opposite sides are equal. Therefore, $CD = AB = 4\text{ cm}$ and $AD = BC = 5\text{ cm}$.

5. With $C$ as center and radius $4\text{ cm}$, draw an arc.

6. With $A$ as center and radius $5\text{ cm}$, draw another arc intersecting the previous arc at point $D$.

7. Join $AD$ and $CD$.

8. $ABCD$ is the required parallelogram.

Given:

In rhombus $ABCD$, side length $= 5\text{ cm}$ and $\angle A = 60^\circ$.

To Construct:

A rhombus $ABCD$.

Steps of Construction:

1. Draw a line segment $AB = 5\text{ cm}$.

2. At point $A$, construct an angle $\angle XAB = 60^\circ$.

3. With $A$ as center and radius $5\text{ cm}$, draw an arc on ray $AX$ to mark point $D$.

4. In a rhombus, all sides are equal. Thus, $BC = 5\text{ cm}$ and $CD = 5\text{ cm}$.

5. With $D$ as center and radius $5\text{ cm}$, draw an arc.

6. With $B$ as center and radius $5\text{ cm}$, draw another arc intersecting the previous arc at point $C$.

7. Join $BC$ and $CD$.

8. $ABCD$ is the required rhombus.

Given:

In rectangle $ABCD$, one side (let $AB$) $= 3\text{ cm}$ and diagonal (let $AC$) $= 5\text{ cm}$.

To Construct:

A rectangle $ABCD$.

Steps of Construction:

1. Draw a line segment $AB = 3\text{ cm}$.

2. At point $B$, construct an angle of $90^\circ$ (since all angles of a rectangle are $90^\circ$). Let the ray be $BY$.

3. With $A$ as center and radius $5\text{ cm}$ (length of diagonal), draw an arc intersecting ray $BY$ at point $C$.

4. Measure the length $BC$. (By Pythagoras theorem, $BC = \sqrt{5^2 - 3^2} = 4\text{ cm}$).

5. With $A$ as center and radius equal to $BC$ ($4\text{ cm}$), draw an arc.

6. With $C$ as center and radius equal to $AB$ ($3\text{ cm}$), draw another arc intersecting the previous arc at point $D$.

7. Join $AD$ and $CD$.

8. $ABCD$ is the required rectangle.

Given:

In square $ABCD$, side length $= 4\text{ cm}$.

To Construct:

A square $ABCD$.

Steps of Construction:

1. Draw a line segment $AB = 4\text{ cm}$.

2. At point $A$, construct an angle of $90^\circ$ and draw ray $AX$.

3. At point $B$, construct an angle of $90^\circ$ and draw ray $BY$.

4. With $A$ as center and radius $4\text{ cm}$, draw an arc on ray $AX$ to mark point $D$.

5. With $B$ as center and radius $4\text{ cm}$, draw an arc on ray $BY$ to mark point $C$.

6. Join $CD$.

7. $ABCD$ is the required square where all sides are $4\text{ cm}$ and all angles are $90^\circ$.

Given:

In rhombus $CLUE$:

1. Side $CL = 7.5\text{ cm}$. Since it is a rhombus, all sides are equal ($CL = LU = UE = EC = 7.5\text{ cm}$).

2. Diagonal $LE = 6\text{ cm}$.

Steps of Construction:

1. Draw a line segment $CL = 7.5\text{ cm}$.

2. With $C$ as center and radius $7.5\text{ cm}$, draw an arc.

3. With $L$ as center and radius $6\text{ cm}$ (diagonal $LE$), draw another arc intersecting the previous arc at point $E$.

4. Join $CE$ and $LE$.

5. With $E$ as center and radius $7.5\text{ cm}$, draw an arc.

6. With $L$ as center and radius $7.5\text{ cm}$, draw an arc intersecting the previous arc at point $U$.

7. Join $EU$ and $LU$.

8. $CLUE$ is the required rhombus.

Given:

In quadrilateral $BEAR$:

$BE = 6\text{ cm}$, $EA = 7\text{ cm}$, $RB = 5\text{ cm}$, $RE = 5\text{ cm}$ and diagonal $BA = 9\text{ cm}$.

Steps of Construction:

1. Draw a line segment $BE = 6\text{ cm}$.

2. With $B$ as center and radius $5\text{ cm}$, and with $E$ as center and radius $5\text{ cm}$, draw arcs that intersect at point $R$. Join $BR$ and $ER$.

3. Now, with $B$ as center and radius $9\text{ cm}$ (diagonal $BA$), draw an arc.

4. With $E$ as center and radius $7\text{ cm}$, draw another arc intersecting the previous arc at point $A$.

5. Join $EA$ and $BA$.

6. Join $RA$ to complete the quadrilateral $BEAR$.

To Find:

Measure the fourth side $RA$.

On measuring the length of $RA$ using a scale, we find:

$RA \approx 4.8\text{ cm}$ (The exact value may vary slightly depending on construction precision).

Given:

In parallelogram $POUR$:

$PO = 5.5\text{ cm}$, $OU = 7.2\text{ cm}$, and $\angle O = 70^\circ$.

Steps of Construction:

1. Draw a line segment $OU = 7.2\text{ cm}$.

2. At point $O$, construct an angle $\angle XOU = 70^\circ$ using a protractor.

3. With $O$ as center and radius $5.5\text{ cm}$, draw an arc on ray $OX$ to mark point $P$.

4. Since opposite sides of a parallelogram are equal, $PR = OU = 7.2\text{ cm}$ and $UR = PO = 5.5\text{ cm}$.

5. With $P$ as center and radius $7.2\text{ cm}$, draw an arc.

6. With $U$ as center and radius $5.5\text{ cm}$, draw an arc intersecting the previous arc at point $R$.

7. Join $PR$ and $UR$.

8. $POUR$ is the required parallelogram.

Given:

A circle with radius $r = 3\text{ cm}$. Diameter $AC = 2r = 6\text{ cm}$.

Steps of Construction:

1. Mark a point $O$ as center and draw a circle of radius $3\text{ cm}$.

2. Draw a diameter and label its endpoints as $A$ and $C$.

3. Construct the perpendicular bisector of $AC$. This line will pass through the center $O$.

4. Let this perpendicular bisector intersect the circle at points $B$ and $D$.

5. Join $AB, BC, CD,$ and $DA$.

Finding the type of Quadrilateral:

The quadrilateral $ABCD$ is a Square.

Justification:

1. $AC$ and $BD$ are both diameters of the circle. Therefore, $AC = BD = 6\text{ cm}$. (Diagonals are equal).

2. Both diameters intersect at the center $O$, so they bisect each other ($AO = OC = BO = OD = 3\text{ cm}$). (Diagonals bisect each other).

3. By construction, $BD$ is the perpendicular bisector of $AC$, so $\angle AOB = 90^\circ$. (Diagonals are perpendicular).

A quadrilateral whose diagonals are equal, bisect each other, and are perpendicular to each other is a Square.

Given:

In parallelogram $HOME$:

Adjacent sides $HO = 6\text{ cm}$ and $HE = 4\text{ cm}$. Diagonal $OE = 3\text{ cm}$.

Steps of Construction:

1. Draw a line segment $HO = 6\text{ cm}$.

2. With $H$ as center and radius $4\text{ cm}$ (side $HE$), draw an arc.

3. With $O$ as center and radius $3\text{ cm}$ (diagonal $OE$), draw an arc intersecting the previous arc at point $E$.

4. Join $HE$ and $OE$.

5. Since opposite sides of a parallelogram are equal, $EM = HO = 6\text{ cm}$ and $OM = HE = 4\text{ cm}$.

6. With $E$ as center and radius $6\text{ cm}$, draw an arc.

7. With $O$ as center and radius $4\text{ cm}$, draw an arc intersecting the previous arc at point $M$.

8. Join $EM$ and $OM$.

9. $HOME$ is the required parallelogram.

Given:

$AB = 3\text{ cm}$, $BC = 4\text{ cm}$, $CD = 5.4\text{ cm}$, $DA = 5.9\text{ cm}$ and diagonal $AC = 8\text{ cm}$.

Solution:

To construct a quadrilateral, it must be possible to construct the two triangles it is composed of, which are separated by the diagonal.

Let's check the constructibility of $\triangle ABC$ using the Triangle Inequality Property (the sum of any two sides of a triangle must be greater than the third side).

In $\triangle ABC$:

Side $AB = 3\text{ cm}$

Side $BC = 4\text{ cm}$

Side $AC = 8\text{ cm}$

Now, let's find the sum of the two smaller sides:

$AB + BC = 3\text{ cm} + 4\text{ cm} = 7\text{ cm}$

Here, we observe that:

Since the sum of two sides is less than the third side, $\triangle ABC$ cannot be formed. If the triangle cannot be formed, the quadrilateral cannot be constructed.

Final Answer: No, it is not possible to construct the quadrilateral $ABCD$ because the triangle inequality property is not satisfied for $\triangle ABC$.

Given:

In quadrilateral $ROAM$:

$\angle O = 120^\circ$, $\angle R = 105^\circ$ and $\angle A = 135^\circ$.

Solution:

We know that according to the Angle Sum Property of a quadrilateral, the sum of all four interior angles must be exactly $360^\circ$.

Let's calculate the sum of the three given angles:

Sum $= \angle R + \angle O + \angle A$

Sum $= 105^\circ + 120^\circ + 135^\circ$

Sum $= 360^\circ$

The sum of just three angles of the quadrilateral is already $360^\circ$. This implies that the fourth angle ($\angle M$) would have to be:

$\angle M = 360^\circ - 360^\circ = 0^\circ$

A quadrilateral cannot have an angle of $0^\circ$ as it would then cease to be a four-sided closed figure (it would collapse into a triangle or a line).

Final Answer: No, it is not possible to construct this quadrilateral because the sum of the given three angles is $360^\circ$, leaving no room for a fourth angle.

Given:

In a square, both diagonals are equal and they bisect each other at right angles ($90^\circ$).

Diagonal length $= 5\text{ cm}$.

Steps of Construction:

1. Draw a line segment $AC = 5\text{ cm}$ (this is the first diagonal).

2. Draw the perpendicular bisector of $AC$. Let it intersect $AC$ at point $O$. $O$ is the midpoint of $AC$.

3. Since the diagonals bisect each other, the distance from the center $O$ to each vertex must be $\frac{5}{2} = 2.5\text{ cm}$.

4. With $O$ as center and radius $2.5\text{ cm}$, draw two arcs on the perpendicular bisector line, one above $AC$ and one below $AC$. Mark these points as $B$ and $D$.

5. Join $AB, BC, CD,$ and $DA$.

6. $ABCD$ is the required square.

Given:

$NE = 7\text{ cm}$, $EW = 6\text{ cm}$, $\angle N = 60^\circ$, $\angle E = 110^\circ$ and $\angle S = 85^\circ$.

Solution:

Before construction, we need the fourth angle $\angle W$ to proceed from the side $EW$.

Using Angle Sum Property of a quadrilateral:

$\angle N + \angle E + \angle W + \angle S = 360^\circ$

$60^\circ + 110^\circ + \angle W + 85^\circ = 360^\circ$

$255^\circ + \angle W = 360^\circ$

$\angle W = 360^\circ - 255^\circ = 105^\circ$

Steps of Construction:

1. Draw a line segment $NE = 7\text{ cm}$.

2. At point $E$, construct an angle $\angle NEX = 110^\circ$ using a protractor.

3. With $E$ as center and radius $6\text{ cm}$, draw an arc on ray $EX$ to mark point $W$.

4. At point $W$, construct an angle $\angle EWY = 105^\circ$.

5. At point $N$, construct an angle $\angle ENZ = 60^\circ$.

6. Let the rays $WY$ and $NZ$ intersect at point $S$.

7. $NEWS$ is the required quadrilateral.

Given:

In parallelogram $ABCD$:

Side $AB = 4\text{ cm}$.

Diagonal $AC = 7\text{ cm} \implies$ Half diagonal $OA = 3.5\text{ cm}$.

Diagonal $BD = 5.6\text{ cm} \implies$ Half diagonal $OB = 2.8\text{ cm}$.

Steps of Construction:

1. Draw the side $AB = 4\text{ cm}$.

2. With $A$ as center and radius $3.5\text{ cm}$, draw an arc.

3. With $B$ as center and radius $2.8\text{ cm}$, draw another arc intersecting the previous arc at point $O$. ($O$ is the intersection of diagonals).

4. Join $AO$ and produce it to point $C$ such that $OC = OA = 3.5\text{ cm}$.

5. Join $BO$ and produce it to point $D$ such that $OD = OB = 2.8\text{ cm}$.

6. Join $BC, CD,$ and $DA$.

7. $ABCD$ is the required parallelogram.

To Find:

Measure the other side $BC$ (or $AD$).

By using a scale to measure $BC$ on the constructed figure, we find:

$BC \approx 4.5\text{ cm}$ (Value may vary slightly with construction).

Given:

Number of sides of the regular polygon, $n = 20$.

To Find:

The measure of each interior angle.

Solution:

We know that the formula for the measure of each interior angle of a regular polygon with $n$ sides is:

$\text{Each Interior Angle} = \frac{(n - 2) \times 180^\circ}{n}$

Substituting the value of $n = 20$ in the formula:

$\text{Each Interior Angle} = \frac{(20 - 2) \times 180^\circ}{20}$

$\text{Each Interior Angle} = \frac{18 \times 180^\circ}{20}$

$\text{Each Interior Angle} = 18 \times \frac{\cancel{180}^{9}}{\cancel{20}_{1}}$

$\text{Each Interior Angle} = 18 \times 9^\circ$

$\text{Each Interior Angle} = 162^\circ$

Final Answer: The measure of each angle of a regular polygon with 20 sides is 162°.

Given:

In trapezium $RISK$, parallel sides are $RI$ and $KS$ ($RI \parallel KS$).

$RI = 7\text{ cm}$, $IS = 5\text{ cm}$, $RK = 6.5\text{ cm}$ and $\angle I = 60^\circ$.

To Construct:

A trapezium $RISK$.

Steps of Construction:

1. Draw a line segment $RI = 7\text{ cm}$.

2. At point $I$, construct an angle $\angle XIR = 60^\circ$ using a protractor or compass.

3. With $I$ as center and radius $5\text{ cm}$, draw an arc on ray $IX$ to mark point $S$. Thus, $IS = 5\text{ cm}$.

4. Since $RI \parallel KS$, at point $S$, construct a line $SY$ parallel to $RI$. This can be done by constructing an angle $\angle YSI = 120^\circ$ (since co-interior angles $60^\circ + 120^\circ = 180^\circ$ for parallel lines).

5. With $R$ as center and radius $6.5\text{ cm}$, draw an arc intersecting the ray $SY$ at point $K$.

6. Join $RK$ and $SK$.

7. $RISK$ is the required trapezium.

Given:

In trapezium $ABCD$, $AB \parallel CD$, $AD = BC = 3.2\text{ cm}$, $AB = 6.4\text{ cm}$ and $CD = 9.6\text{ cm}$.

Construction Required:

On side $CD$, mark a point $E$ such that $DE = AB = 6.4\text{ cm}$. Join $BE$.

Steps of Construction:

1. Draw a line segment $CD = 9.6\text{ cm}$.

2. Mark point $E$ on $CD$ such that $DE = 6.4\text{ cm}$. Then $EC = CD - DE = 9.6\text{ cm} - 6.4\text{ cm} = 3.2\text{ cm}$.

3. Construct $\triangle BEC$ such that $BC = 3.2\text{ cm}$ and $BE = 3.2\text{ cm}$ (as $BE \parallel AD$ and $BE = AD$).

4. Since $BC = EC = BE = 3.2\text{ cm}$, $\triangle BEC$ is an equilateral triangle.

5. From point $B$, draw a line $BA$ parallel to $ED$ such that $BA = 6.4\text{ cm}$.

6. Join $AD$. $ABCD$ is the required trapezium.

Solution:

In equilateral $\triangle BEC$:

Since $AD \parallel BE$ and $CD$ is the transversal:

In trapezium $ABCD$, since $AB \parallel CD$:

$\angle A + 60^\circ = 180^\circ$

$\angle A = 120^\circ$

Similarly, for parallel lines $AB$ and $CD$ with transversal $BC$:

$\angle B + 60^\circ = 180^\circ$

$\angle B = 120^\circ$

Final Answer: On measuring the angles, the values are $\angle A = 120^\circ$ and $\angle B = 120^\circ$.